Prove that if $X$ and $Y$ are $\in$–isomorphic transitive sets, then $X = Y$ and the identity on $X$ is the unique $\in$–isomorphism $f\colon (X, \in) \to (Y, \in)$.
I've started this question, and hoped to prove it by assuming that X doesn't equal Y for a contradiction, but i'm not sure where to go from this, or if this is even the right approach.
Thanks for any help.
Whenever well-founded objects are involved, the first word that should come to mind is "induction".
So we can prove this by induction on the rank of members of $X$. The goal, of course, is to prove that $f(x)=x$ for all $x\in X$.
Now suppose this is true for all the elements of $x$. Namely, if $x'\in x$, then $f(x')=x'$. Let $y$ denote $f(x)$.
$x\subseteq y$, since for every $x'\in x$, $x'=f(x')\in f(x)=y$.
$y\subseteq x$ holds, since if $y'\in y$, by the fact $f$ is a bijection there is some $x'\in X$ such that $f(x')=y'$, and by the fact it is an isomorphism $x'\in x$.
But the induction hypothesis was that $f(x')=x'$, since $x'\in x$. Therefore $y\subseteq x$.
From the fact $f$ is the identity we get that $X=Y$.
Note that we rely quite heavily on the axiom of regularity, but that's for a reason. Suppose it fails and $x=\{x\}\neq \{y\}=y$, then both $x$ and $y$ are transitive and isomorphic, but the isomorphism is not the identity.