Why do we care about two subgroups being conjugate?

In classifications of the subgroups of a given group, results are often stated up to conjugacy. I would like to know why this is.

More generally, I don't understand why "conjugacy" is an equivalence relation we care about, beyond the fact that it is stronger than "abstractly isomorphic."

My vague understanding is that while "abstractly isomorphic" is the correct "intrinsic" notion of isomorphism, so "conjugate" is the correct "extrinsic" notion. But why have we designated this notion of equivalence, and not some other one?

To receive a satisfactory answer, let me be slightly more precise:

Question: Given two subgroups $H_1, H_2$ of a given group $G$, what properties are preserved under conjugacy that may break under general abstract isomorphism?

For example, is it true that $G/H_1 \cong G/H_2$ iff $H_1$ is conjugate to $H_2$? Or, is it true that two subgroups $H_1, H_2 \leq \text{GL}(V)$ are conjugate iff their representations are isomorphic? I'm sure these are easy questions to answer -- admittedly, I haven't thought fully about either -- but I raise them by way of example. What are other such equivalent characterizations?

This answer isn't of the form you asked for, but I'm posting it because it's been very helpful to me. I apologize if this is the sort of answer you're trying to avoid.

Let's start with a very specific situation. Let $G$ be the isometry group of the plane $E$. If you pick a point $x$, the rotations around $x$ form a subgroup $G_x$ of $G$. If you pick two different points $x$ and $y$, you get two subgroups $G_x$ and $G_y$. These subgroups are different, but they don't feel really different, because you can turn one into the other just by changing your point of view: if you shift the plane by an isometry that puts $x$ where $y$ used to be, then $G_x$ becomes the subgroup $G_y$ used to be. In other words, $G_x$ and $G_y$ aren't really different because there's an isometry $g \in G$ that makes the diagram $$\require{AMScd} \begin{CD} E @>G_x>> E \\ @VgVV @VVgV \\ E @>>G_y> E \end{CD}$$ commute. (In this diagram, each arrow stands for a whole set of isometries, with $E \overset{g}{\longrightarrow} E$ denoting the singleton $\{g\}$, and composing arrows means composing all pairs of isometries.)

Now, to be completely general, think of any group $G$ as a group of "allowed symmetries" of some object—in other words, a subgroup of $\operatorname{Aut} X$ for some object $X$. (This is completely general because we can take $X$ to be $G$ itself with the left multiplication action. Perhaps more satisfyingly, we can make $X$ a graph if $G$ is finite, a tree if $G$ is free, an effective Klein geometry if $G$ is a suitable Lie group...)

Once again, even if two subgroups of $G$ are different, they don't feel really different if you can turn one into the other just by looking at $X$ from a different point of view. In other words, two subgroups $H, \tilde{H} < G$ aren't really different if there's an allowed symmetry $g \in G$ of $X$ that makes the diagram $$\require{AMScd} \begin{CD} X @>H>> X \\ @VgVV @VVgV \\ X @>>\tilde{H}> X \end{CD}$$ commute. This, of course, is the definition of conjugacy.

This point of view, by the way, is my favorite motivation for the idea of normality: a normal subgroup is a class of symmetries that doesn't depend on your point of view. If you shift the plane by an isometry, for example, your notion of what counts as a horizontal translation might change, but your notion of what counts as a translation won't. Horizontal translations form a non-normal subgroup, while translations form a normal subgroup.

$H_1$ and $H_2$ are conjugate as subgroups iff $G/H_1$ and $G/H_2$ are isomorphic as $G$-sets.

Edit: Two related settings where this condition shows up are Galois theory and covering space theory. One way to state the classification of (not necessarily connected) covering spaces of a (nice, path-connected) space $X$ is that the category of such covers is equivalent to the category of $\pi_1(X)$-sets. Among these, the transitive $\pi_1(X)$-sets correspond to the connected covers, so we get that connected covers correspond to conjugacy classes of subgroups of $\pi_1(X)$. To get subgroups on the nose you need to pick basepoints in the covers lifting a basepoint in $X$.

Edit #2: Here is another setting where this condition appears, to whet your appetite. If $H_1$ and $H_2$ are isomorphic, then their categories $\text{Rep}(H_1)$ and $\text{Rep}(H_2)$ of linear representations are equivalent. But if $H_1$ and $H_2$ are conjugate, it's furthermore true that we can choose an equivalence between these categories to have the property that the corresponding induction functors $\text{Ind}_{H_i}^G : \text{Rep}(H_i) \to \text{Rep}(G)$ are naturally isomorphic.