When will two isomorphic Lie algebras have the same representation?

Short answer: If two Lie algebras are isomorphic, they have "the same" complex representations. A real semisimple Lie algebra and its complexification also have "the same" complex representations, but only to a certain extent: During complexification, the correspondence "forgets" which representations of the real Lie algebra were conjugate to which. Hence, if two real semisimple Lie algebras happen to have isomorphic complexifications, but are not isomorphic themselves, then there is a one-to-one correspondence between their representations (going "up" from one to their common complexification and then back "down" to the other), but this correspondence has no reason to, and in general does not, respect conjugacy among the representations.

Long answer

If two Lie algebras are isomorphic, there is an obvious bijection between their representations, which preserves dimension, irreducibility, duality, conjugation relations and whatnot. _{(I think the precise math terminology would be that an isomorphism of Lie algebras $\mathfrak{g} \simeq\mathfrak{h}$ induces an equivalence of (abelian, Tannakian?) categories between their respective categories of representations $\mathfrak{g}-Rep \simeq \mathfrak{h}-Rep$ which commutes with blah blah; "bijection" might not be the right word for set-theoretic issues; for practical purposes, let's not go down that rabbit hole and just say $\mathfrak{g}$ and $\mathfrak{h}$ have "the same" representations. It is worth noting that this "equivalence of categories" is on a different level than the usual "equivalence" between two representations of one single Lie algebra, which is the only way the word "equivalence" will be used in the rest of this answer, except in the "Final Footnote".)}

The problem here seems to be that people are sloppy in keeping track of what is a representation of either the complexified Lie algebras or the original real Lie algebras.

Here, the representations of the complex Lie algebras $\mathfrak{so}(3,1)_{\mathbb C}$ and $\mathfrak{so}(4)_{\mathbb C}$ are "the same" because these Lie algebras are $\mathbb C$-isomorphic; in particular, up to equivalence there are two irreducible ones on $2$-dimensional $\mathbb C$-vector spaces, let's call them $\rho_1$ and $\rho_2$.

But what the sources actually talk about are the restrictions of these representations to the real subalgebras $\mathfrak{so}(3,1)$ resp. $\mathfrak{so}(4)$ sitting inside the complexified one; and these restricted representations of real Lie algebras (still on $\mathbb C^2$ though) can behave differently. Let's have a closer look.

Let us fix the big complexified Lie algebra in which everything lives as $\mathfrak{sl}_2(\mathbb C) \oplus \mathfrak{sl}_2(\mathbb C)$ once and for all. Written like this, we can say $\rho_1$ is the natural action of the first summand on $\mathbb C^2$, and $\rho_2$ is the natural action of the the second summand on $\mathbb C^2$.

Now how do the real Lie algebras $\mathfrak{so}(4)$ and $\mathfrak{so}(3,1)$ sit inside that direct sum? The Wikipedia article you quote gives (in the section "The Lie algebra") certain elements $A_j$ and $B_j$ $(j=1,2,3)$, so that the $A_j$ form a complex basis of the first summand $\mathfrak{sl}_2(\mathbb C)$, and the $B_j$ a complex basis for the second summand $\mathfrak{sl}_2(\mathbb C)$. However, each of the triples $(A_j)_j$ resp. $(B_j)_j$ also forms a basis of a real Lie algebra $\mathfrak{su}(2)$; so if we look at the real Lie algebra with all six of these elements as basis, we get $\mathfrak{su}(2)\oplus \mathfrak{su}(2)$, and that is our real subalgebra $\mathfrak{so}(4)$:

$\mathfrak{so}(4) = \mathfrak{su}(2) \oplus \mathfrak{su}(2) \subset \mathfrak{sl}_2(\mathbb C) \oplus \mathfrak{sl}_2(\mathbb C)$.

Now if we look at

$\rho_{1 \vert \mathfrak{so}(4)}$ and $\rho_{2 \vert \mathfrak{so}(4)}$

we see that $\rho_{1 \vert \mathfrak{so}(4)}$ is just the fundamental quaternionic representation of the first summand $\mathfrak{su}(2)$ (i.e. $\mathfrak{su}(2)$ acting on $\mathbb C^2$), and $\rho_{2 \vert \mathfrak{so}(4)}$ the fundamental one of the second one. Each of them is indeed self-conjugate and pseudoreal (quaternionic).

On the other hand, the real Lie algebra $\mathfrak{so}(3,1)$ sits "skew" in the direct sum $\mathfrak{sl}_2(\mathbb C) \oplus \mathfrak{sl}_2(\mathbb C)$. Indeed, following the conventions in the Wikipedia article, a basis of this is given by the six elements $J_j = A_j + B_j $ and $K_j = (-i) \cdot (A_j - B_j)$, $j=1,2,3$:

$\mathfrak{so}(3,1) = \sum \mathbb R (A_j + B_j) + \sum (-i)\mathbb R(A_j-B_j) \subset \mathfrak{sl}_2(\mathbb C) \oplus \mathfrak{sl}_2(\mathbb C)$.

So if, now, we look at the restrictions

$\rho_{1 \vert \mathfrak{so}(3,1)}$ and $\rho_{2 \vert \mathfrak{so}(3,1)}$

then we see that the first one sends (for $j=1,2,3$) $J_j$ to $A_j$ and $K_j$ to $\color{red}{-i} A_j$ (acting on $\mathbb C^2$); whereas the second one sends $J_j$ to $B_j$, but $K_j$ to $\color{red}{i} B_j$ (acting on $\mathbb C^2$). Now replacing the $A_j$ and $B_j$ by the standard normed Pauli matrices, which according to the article satisfy exactly the commutation relation which the $A$- resp. $B$-triples do, this gives exactly the representations you describe, and yes, they are related to each other by conjugation.

Related questions: It has already confused many people that the Lie algebra called $\mathfrak{so}(3,1)$ here is also isomorphic to $\mathfrak{sl}_2(\mathbb C)_{\mathbb R}$, i.e. the complex Lie algebra $\mathfrak{sl}_2(\mathbb C)$ viewed as a six-dimensional real Lie algebra. Cf. Precise connection between complexification of $\mathfrak{su}(2)$, $\mathfrak{so}(1,3)$ and $\mathfrak{sl}(2, \mathbb{C})$, Relationship between proper orthochronous Lorentz group $SO^+(1,3)$ and $SU(2)\times SU(2)$, or their Lie algebras, Representations of $sl(2,C)$ as a real Lie algebra, https://physics.stackexchange.com/q/108212/168529.

Further, the two inequivalent, but conjugate to each other, irreps called $\rho_{1 \vert \mathfrak{so}(3,1)}$ and $\rho_{2 \vert \mathfrak{so}(3,1)}$ here have featured in Conjugate Representations for $\mathfrak{sl}(2,\mathbb{C})$, Why are the fundamental and anti-fundamental representation in $\text{SL}(2,\mathbb{C})$ not equivalent?, and Conjugate Representations of Lie Algebra of Lorentz Group. The answers to the first two (by the user Qmechanic, whose answers to related posts here and on physics.stackexchange have helped my understanding a lot) swiftly clear some confusion (in particular by insisting that "conjugate" only makes sense for representations restricted to something real), whereas the answer to the third one in my eyes is incomplete, as it again confuses representations of the complexified algebra with their restrictions to $\mathfrak{so}(3,1)$.

A slightly different example

Let's go through another example and write everything down with matrices to see things clearer.

Let's look at the complex Lie algebra $\mathfrak{sl}_3(\mathbb C)$. It is simple, of complex dimension $8$, and its representation theory with weights and stuff is treated extensively in the literature. In particular, it has two irreps on complex vector spaces of dimension $3$, namely

the "standard" or "defining" representation

$A: \mathfrak{sl}_3(\mathbb C) \rightarrow End_{\mathbb C}(\mathbb C^3)$, where

$X$ acts on $v \in \mathbb C^3$ by normal matrix multiplication $X\cdot v$, i.e. $A(X) = X$;

then there is its dual representation

$B: \mathfrak{sl}_3(\mathbb C) \rightarrow End_{\mathbb C}(\mathbb C^3)$, where

$X$ acts on $v \in \mathbb C^3$ by multiplication with its negative transpose $-X^{tr}\cdot v$, i.e. $B(X) = -X^{tr}$.

Many mathematicians would call both of them "fundamental representations" (they belong to the two fundamental weights of the root system respectively), whereas in physics literature it seems common to call $A$ "the fundamental" representation and $B$ the "antifundamental" one. _{(To see these different terminologies clash, cf. e.g. What are defining & fundamental representations?, Number of non-equivalent fundamental representations for a Lie algebra, Fundamental, Regular, and Defining representations.)}

It is standard to show that $A$ and $B$ are not equivalent to each other. Another common notation is "$\mathbf 3$" for $A$ and "$\bar{\mathbf 3}$" for $B$, although these, because of the overline suggesting complex conjugation, should be saved for certain restricted representations we will look at now.

Namely, there are three real Lie algebras

$\mathfrak{g}_1 = \mathfrak{sl}_3(\mathbb R) = \lbrace \begin{pmatrix} a & c & e\\ f & b & d\\ h & g & -a-b \end{pmatrix} : a, ..., h \in \mathbb{R} \rbrace$;

$\mathfrak{g}_2 = \mathfrak{su}_{1,2} := \lbrace \begin{pmatrix} a+bi & c+di & ei\\ f+gi & -2bi & -c+di\\ hi & -f+gi & -a+bi \end{pmatrix} : a, ..., h \in \mathbb{R} \rbrace$;

$\mathfrak{g}_3 = \mathfrak{su}_{3} := \lbrace \begin{pmatrix} ia & c+di & g+hi\\ -c+di & ib & e+fi\\ -g+hi & -e+fi & -ai-bi \end{pmatrix} : a, ..., h \in \mathbb{R} \rbrace$.

which written this way all "sit inside" our complex Lie algebra $\mathfrak{sl}_3(\mathbb C)$. Each of them is simple and $8$-(real-)dimensional, and extending scalars we see that they all have complexification $(\mathfrak{g}_i)_\mathbb C = \mathbb C \otimes_\mathbb R \mathfrak{g}_i \simeq \mathfrak{sl}_3(\mathbb C)$, meaning that they are "real forms" of $\mathfrak{sl}_3(\mathbb C)$. As real Lie algebras, they are mutually non-isomorphic though; it turns out that these three are, up to isomorphism, all the real forms of $\mathfrak{sl}_3(\mathbb C)$, i.e. all the simple real Lie algebras which have complexification $\mathfrak{sl}_3(\mathbb C)$. The first one is called "the split form", the second one here can be called "the quasi-split form", and the third one is "the compact form".

Now what are interesting representations of $\mathfrak{g}_{1,2,3}$? And indeed, mathematicians and especially physicists here almost always mean representations on complex vector spaces, even though the Lie algebras are real; i.e. when talking of three-dimensional representations, we still mean Lie algebra homomorphisms

$$\mathfrak{g}_i \rightarrow End_{\mathbb C}(\mathbb C^3)$$

(the map must be real-linear, and cannot be more, since on the LHS only real scalars act).

But we defined all the $\mathfrak{g}_i$ as subsets of $\mathfrak{sl}_3(\mathbb C)$, and we have those two maps

$A, B: \mathfrak{sl}_3(\mathbb C) \rightarrow End_{\mathbb C}(\mathbb C^3)$

so let’s just restrict them to each of the $\mathfrak{g}_i$ and call the result $A_i$ resp. $B_i$. So for example

$A_1$ is just the map $\mathfrak{g}_1 \rightarrow End_{\mathbb C}(\mathbb C^3)$

that lets $\begin{pmatrix} a & c & e\\ f & b & d\\ h & g & -a-b \end{pmatrix}$ ($a, ..., h \in \mathbb{R}$) act on $v \in \mathbb C^3$ via $\begin{pmatrix} a & c & e\\ f & b & d\\ h & g & -a-b \end{pmatrix} \cdot v$;

whereas

$B_2$ is the map $\mathfrak{g}_2 \rightarrow End_{\mathbb C}(\mathbb C^3)$

that lets $\begin{pmatrix} a+bi & c+di & ei\\ f+gi & -2bi & -c+di\\ hi & -f+gi & -a+bi \end{pmatrix}$ ($a, ..., h \in \mathbb R$) act on $v \in \mathbb C^3$ via $\begin{pmatrix} -a-bi & -f-gi & -hi\\ -c-di & 2bi & f-gi\\ -ei & c-di & a-bi \end{pmatrix} \cdot v$

etc.

For each $i$, $A_i$ and $B_i$ are two irreducible complex representations of $\mathfrak{g}_i$, inequivalent to each other; and there can be no further inequivalent ones. This is maybe not as trivial as it sounds, but let's accept it for now. _{(See "Final Footnote".)}

Now we got real Lie algebras acting via complex matrices on complex vector spaces, so we can start looking at conjugate representations. Namely, for each of the maps $A_i, B_i$, we can just complex-conjugate the matrix $A_i(X)$ resp. $B_i(X)$, and in each case this will give (seemingly) new representations

$\overline{A_i}, \overline{B_i}: \mathfrak{g}_i \rightarrow End_{\mathbb C}(\mathbb C^3)$

But let’s see what actually happens. First, the split case $i=1$. Well the matrices $A_1(X)$ and $B_1(X)$ for $X \in \mathfrak{g}_1$ have all real entries, so conjugation literally does nothing, and we have

$$\overline{A_1} = A_1, \overline{B_1} = B_1.$$

On the other extreme $i=3$, looking at the matrices that make up the compact form $\mathfrak{g}_3$, we see that on those, complex conjugating is exactly the same as taking the negative transpose! Hence

$$\overline{A_3} = B_3,$$ $$\overline{B_3} = A_3.$$

(Note that the "big" complex representations $A$ and $B$ were dual to each other; here we see a special case of the fact that when restricting to the "unitary" $\mathfrak{g}_3= \mathfrak{su}_3$, the conjugate representation "is" the dual representation, or more precisely: the conjugate of the restriction is the restriction of the dual). _{(The literature, especially physics literature, seems to use "dual" and "conjugate" almost interchangeably, which because of what I just said might be convenient as long as everyone is aware that we only talk about representations of real unitary groups/algebras on complex vector spaces; but since that's rarely mentioned, it confused the hell out of me. As an aside, in the basic example of $\mathfrak{su}_2$, which physicists and mathematicians agree has only one fundamental representation, this irrep happens to be self-dual and (thus) self-conjugate; funnily here the fact that conjugating "is" dualising i.e. switches the weights / eigenvalues has also confused someone: https://physics.stackexchange.com/q/139532/168529.)}

It are these representations $A_3$ and $B_3$ of the compact form $\mathfrak{su}_3$ which should properly be called "$\mathbf 3$" and "$\overline{\mathbf 3}$", because these are actually conjugate to each other.

Finally, what happens in the quasi-split $i=2$ case in the middle? This is a bit more fishy. E.g.

$\overline{A_2}$ is the map $\mathfrak{g}_2 \rightarrow End_{\mathbb C}(\mathbb C^3)$

that lets $\begin{pmatrix} a+bi & c+di & ei\\ f+gi & -2bi & -c+di\\ hi & -f+gi & -a+bi \end{pmatrix}$ ($a, ..., h \in \mathbb{R}$) act on $v \in \mathbb C^3$ via $\begin{pmatrix} a-bi & c-di & -ei\\ f-gi & 2bi & -c-di\\ -hi & -f-gi & -a-bi \end{pmatrix} \cdot v$

and on first sight, this is different from both $A_2$ and $B_2$. So does $\mathfrak{g}_2$ have more irreps? No. Check for yourself that

$$\overline{A_2} = \begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0 \end{pmatrix}^{-1} \cdot B_2 \cdot \begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0 \end{pmatrix}$$

meaning that $\overline{A_2}$ is equivalent to $B_2$, and conversely, $\overline{B_2} \simeq A_2$.

Summing up:

• The common complexification $\mathfrak{sl}_3(\mathbb C)$ has two inequivalent irreps on $\mathbb C^3$, which are dual to each other, $A$ and $B$.
• Each of the real Lie algebras $\mathfrak{g}_i$ has two inequivalent irreps on $\mathbb C^3$:
• for the split form, each of these two ($A_1$ and $B_1$) is self-conjugate;
• for the quasi-split form, the conjugate of $A_2$ is equivalent to $B_2$, and vice versa;
• for the compact form, the conjugate of $A_3$ "is" $B_3$, and vice versa.

Note: In your example, $\mathfrak{so}(3,1)$ resp. $\mathfrak{so}(4)$ are a quasi-split resp. compact form of $\mathfrak{sl}_2(\mathbb C) \oplus \mathfrak{sl}_2(\mathbb C)$; you can add the split real form $\mathfrak{sl}_2(\mathbb R) \oplus \mathfrak{sl}_2(\mathbb R)$ to the consideration and see how the restrictions to this one behave under conjugation. Also note that in that example contrary to this one, the representations I called $\rho_1$ and $\rho_2$ are not dual to each other, rather each of them is self-dual, which, together with my boldface point about conjugate vs. dual here, explains why their respective restrictions to the real unitary Lie algebra are self-conjugate.

Final Footnote: After introducing the real forms $\mathfrak{g}_i$ and the restricted representations $A_i, B_i$, I wrote: "For each $i$, $A_i$ and $B_i$ are two irreducible complex representations of $\mathfrak{g}_i$, inequivalent to each other; and there can be no further inequivalent ones. This is maybe not as trivial as it sounds". Indeed I have come to believe that stopping and wondering for a moment about this is the crucial point of this entire answer: For each real form $\mathfrak{g}_i$ of a complex Lie algebra $\mathfrak{g}_{\mathbb C}$, restriction induces a category equivalence between the the (complex!) representations of $\mathfrak{g}_i$ and the (complex!) representations of $\mathfrak{g}_{\mathbb C}$. Cf. e.g. representations of real forms of simple lie algebra, Real vs Complex Representations of a Lie algebra, In what sense are complex representations of a real Lie algebra and complex representations of the complexified Lie algebra equivalent?. This category equivalence, properly understood, respects irreducibility, dimension and other things, however it does not respect "conjugacy" of representations for the striking reason that that notion is not even defined on the side of $\mathfrak{g}_{\mathbb C}-Rep$; and while we saw that for certain compact forms (here, $\mathfrak{g}_3$), conjugacy in $\mathfrak{g}_i-Rep$ translates to duality in $\mathfrak{g}_{\mathbb C}-Rep$, this is generally not true for other real forms. -- And as a footnote to this final footnote, the exclamation mark in "(complex!)" is truly important: There is no restriction/tensoring equivalence, neither from complex $\mathfrak{g}_i-Rep$ nor complex $\mathfrak{g}_{\mathbb C}-Rep$, to representations of either on real vector spaces. This is alluded to in comments to https://math.stackexchange.com/a/1026919/96384 and Obtaining representation of a real Lie algebra from the complexification "by restriction". Relatedly, notice that if one considered the representations $A_1, B_1$ of the split real form $\mathfrak{g}_1$ as representations on the real vector space $\mathbb C^3 \simeq \mathbb R^6$, neither of them would be irreducible any longer; whereas, if I'm not mistaken, e.g. $A_3$ and $B_3$ would still define irreducible representations on $\mathbb R^6$.