First part of the question:
A $G$-set $X$ is transitive if it has only one orbit. Two $G$-sets $X$ and $Y$ are isomorphic if there is a bijection $\phi : X \rightarrow Y$ such that $\phi(gx) = g\phi(x)$ for all $g ∈ G$, $x ∈ X$. Prove a version of the orbit-stabiliser theorem: a transitive $G$-set is isomorphic to $G/A$ where $A$
is a stabiliser of a point.
For the first part, presumably you define a homomorphism with kernel $A$. Instinctively, fixing some point $x \in X$, $\psi : G \rightarrow X$ with $\psi(g) = gx$ whose image will be the orbit of $x$ which is $X$ by assumption, and whose kernel will be $A := Stab(x)$ so we can use the First Isomorphism Theorem for $G/A \cong X$
But $\psi$ is not a homomorphism to $X$, because there's no operation defined in $X$, but I would have though it had to have image $X$ so thaat I could use the First Isomorphism Theorem. Please clarify this if you can, or tell me my method is wrong.
Next part, that may become clear with an answer to the first part, but I don't see how it follows at all:
Prove that the two transitive $G$-sets $G/A$ and $G/B$ are isomorphic if and only if there exists $x ∈ G$ such that $xAx^{−1} = B$.
Best Answer
You can't use isomorphism theorems precisely because $X$ is not a group.
The map, $\psi$, you've given is close to the map you want. Consider instead $\phi:G/A\to X$ defined by $\phi(gA)=gx$ for some $x\in X$ where $A=\mathrm{Stab}(x)$.
This is clearly well-defined and injective as, $gA=hA$ if and only if $g^{-1}h\in A$ if and only if $g^{-1}hx=x$ if and only if $hx=gx$. It's also surjective, because $G$ is transitive so for any $y\in X$ you have $y=gx$ for some $g\in G$ so $\psi(gA)=y$. Finally it is an isomorphism since $\phi(ghA)=ghx=g\phi(hA)$.
For the second part, the stabiliser of $xA$ is $xAx^{-1}$, so $G/A$ is isomorphic to $G/xAx^{-1}$ by the first part. Conversely if $G/A$ is isomorphic to $G/B$ then there is an isomorphism $\psi:G/A\to G/B$. In particular $\phi(xA)=B$ for some $x\in G$. But then $xA=gxA$ if and only if $B=\phi(xA)=\phi(gxA)=g\phi(xA)=gB$, hence $xAx^{-1}=\mathrm{Stab}(xA)=\mathrm{Stab}(B)=B$.