What is an intuitive definition for "conjugate" in Group Theory?

In Abstract Algebra, I learned about "conjugation" in the context of a group $H$ being a 'normal' subgroup of $G$ if the element $xhx^{-1}\in H$ for any $x\in G$. But this is not the first time I've seen the word 'conjugate'. The other times I've seen this are in pre-calculus, when trying to rationalize a denominator, or in the case where $(x+y)$ is the conjugate of $(x-y)$. Does the Group Theory version of conjugate have any link to the pre-calculus version (and other uses)?

C. Falcon's answer refers to Galois theory, which is almost certainly the historically correct answer. But the connection between the two ideas can be explained in slightly more elementary terms:

If $G$ is a group acting on a set $S$, and $s,t\in S$ lie in the same orbit, then the stabilizers of $s$ and $t$ are conjugate subgroups of $G$. Specifically, if $t = hs$, then $gs = s \iff (hgh^{-1})t = t$.

Since we frequently study group elements in terms of their fixed points, it makes sense that the terminology for conjugation in a group gets sometimes mixed up with the terminology for orbits.

For example, we can check that $a+b\sqrt{2} \mapsto a-b\sqrt{2}$ defines an action of the cyclic group of order $2$ on the field $\mathbb{Q}(\sqrt{2})$. Then $a+b\sqrt{2}$ is conjugate to $a-b\sqrt{2}$ in the sense that they lie in the same orbit under this action.

More generally, Galois theory tells us that if $p(X)$ is any irreducible polynomial over a field $k$, and $k\subset K$ is a splitting field, then there is a group acting on $K$ that fixes $k$ and acts transitively on the roots of $p$. In other words, the roots of $p$ are conjugate.

This is somewhat of an abuse of terminology, but it may explain the conceptual link between the two different uses of the word.

Arguably the easiest and the most intuitive way to understand conjugation is through linear algebra.

$T : V \to V$ be a linear transformation of the vector space $V$. Then given a basis $\Bbb B = \{e_1, \cdots, e_n\}$ for $V$, one can represent $T$ by a matrix $[T]_{\Bbb B} = [T(e_1), \cdots, T(e_n)]$, with each $T(e_i)$ expressed in terms of $\Bbb B$. That is, the columns of the matrix are the images of each element of the basis $\Bbb B$ under the transformation.

If $\Bbb B = \{v_1, \cdots, v_n\}$ and $\Bbb B' = \{w_1, \cdots, w_n\}$ are two bases of $V$, then we can try to see what happens if we start with the basis $\Bbb B$ and then switch to the basis $\Bbb B'$ by expressing $v_i$ as linear combination $\sum_k a_{ki} w_i$ of elements of $\Bbb B$. Similarly, we can try expressing elements $w_i$ of $\Bbb B'$ as linear combinations $\sum_k a_{ki}' v_i$ of elements of $\Bbb B$.

The matrix $P = [a_{ki}]$ is called the basechange matrix obtained from changing base from $\Bbb B$ to $\Bbb B'$. It can be easily verified that $[a'_{ki}]$ is simply $P^{-1}$.

The relevant fact is $[T]_{\Bbb B'} = P^{-1} [T]_{\Bbb B} P$. Thus, conjugation of matrices correspond to change of basis and conjugacy classes of matrices in $GL(V)$ correspond to basis-free linear operators.

This description of conjugacy can even be seen in general groups. $G$ be an arbitrary group which acts on a set $X$ by

$$G \times X \to X \\ (g, x) \mapsto gx$$

If $g$ acts on elements on $X$ by multiplication, $hgh^{-1}$ does the same job $g$ does, but on $hx$ instead of $x$, and then puts everything back at place. One can easily see this by letting $\text{Isom}(\Bbb R^3)$ act on $\Bbb R^3$ by isometries, say, and letting $g$ and $h$ to be elements of the isometry group (e.g., rotation/translation).

Thus, $hgh^{-1}$ is not really very different from $g$, but it just does the work of $g$ at $hx$ instead of $x$. This is more general in the sense that letting $G = GL(V)$ and $G$ act on the vector space $X = V$ gives me back the linear algebra picture.

To answer the main question, to my mind, Galois conjugates are not really related to conjugates as in groups (which has already been pointed out by several answers here). The real motivation for conjugation comes from Galois theory in a different way.

Namely, if $\text{Gal}(L/K)$ is the group of $K$-automorphisms of $L$, then given an intermediate extension $L/E/K$ (terminologies are explaned in C. Falcon's answer), $\text{Gal}(L/E)$ is naturally a subgroup of $\text{Gal}(L/K)$, as $E$-automorphisms of $L$ are automatically $K$-automorphisms of $L$ (pointwise fixing something larger automatically pointwise fixes the subset).

Now, if $L/E/K$ is a tower of Galois extensions, then $\text{Gal}(L/E)$ is normal in $\text{Gal}(L/K)$. This is where the notion of conjugacy comes to play, because Galois wanted to define group structure on the collection of all cosets.

If $G$ is a group, $N \leq G$ is a subgroup, it's easy to see where conjugacy comes in trying to define group structure on $G/N$. If $g_1N, g_2N$ are two elements in $G/N$, then the most obvious way to try to define group structure is as $(g_1 N)\cdot(g_2 N) = (g_1g_2)N$. This can be made well-defined if and only if $gNg^{-1} = N$ (one direction is easy, for the other direction if $gN = g'N$, pick $h \in N$. $hN = eN$ implies $(hN) \cdot (gN) = gN$, which is - as we claim - the same as $(hN) \cdot (gN) = (hg)N$. Hence $(g^{-1}hg)N = N$, and $N$ is normal, as desired)

It's interesting as a remark that Galois deduced that the quotient $\text{Gal}(L/K)/\text{Gal}(L/E)$ is in fact isomorphic to $\text{Gal}(K/E)$, although that's irrelevant here.