= Solution
An $\in$-<isomorphism> is a <bijection> $f:X\to Y$ such that $x\in x'$ if and only if $f(x)\in f(x')$ for all $x,x'\in X$. When $Y$ is transitive, <surjectivity> gives the stronger recursive description
$$
f(x)=\{f(u):u\in x\cap X\}\qquad(x\in X).
$$
Indeed every member of $f(x)$ belongs to $Y$ because it is a <transitive set> and so has a preimage in $X$; the <isomorphism> property identifies exactly which preimages occur.
If $f,g$ are two such maps, possibly to different <transitive sets>, form $B=\{x\in X:f(x)\ne g(x)\}$. If $B$ were nonempty, <Foundation> would give $x\in B$ with $x\cap B=\varnothing$. All predecessors $u\in x\cap X$ then satisfy $f(u)=g(u)$, and the recursive descriptions force $f(x)=g(x)$, a contradiction. Therefore \b[the map and its transitive range are unique].
The necessary and sufficient condition is that $\langle X,\in\rangle$ be \b[<well-founded> and extensional as a relation on $X$]. Well-foundedness says every nonempty $A\subseteq X$ has a member with no $\in$-predecessor in $A$. The <extensional relation> condition says distinct elements $x,x'\in X$ have distinct predecessor sets $x\cap X$, $x'\cap X$. Sufficiency is the <Mostowski collapse theorem>.
For necessity, an <isomorphism> transfers well-foundedness from its transitive range, where <Foundation> applies. For extensionality, if $x\cap X=x'\cap X$, the recursive descriptions give $f(x)=f(x')$, and <injectivity> gives $x=x'$. Notice that ambient <axiom of extensionality> alone does not establish this internal predecessor condition; for example $X=\{\{0\},\{2\}\}$ can have distinct elements with no predecessors in $X$.
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