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.
Best Answer
If $\phi : G/H_1 \to G/H_2$ is a homomorphism of $G$-sets, and $\phi([1])=[g]$, then it follows more generally that $\phi([x])=[xg]$. But $\phi$ should be well-defined, i.e. $y^{-1} x \in H_1$ implies $(y g)^{-1} x g \in H_2$. This reduces to $g^{-1} H_1 g \subseteq H_2$. Conversely, this relation implies that $\phi([x]):=[xg]$ is a well-defined homomorphism of $G$-sets. Similarly, $\phi$ is injective iff $g H_2 g^{-1} \subseteq H_1$. And $\phi$ is automatically surjective. It follows that $G/H_1 \cong G/H_2$ as $G$-sets iff $H_1$ and $H_2$ are conjugated.
As already pointed out in the comments, it is really important to work in the category of $G$-sets here. But there aren't any alternatives anyway. Of course the category of sets is too weak, and the category of groups doesn't make sense since $H_1,H_2$ aren't assumed to be normal.