 Solution

An $\in$-isomorphism is a bijection $f:X\to Y$ such that $x\in x^{\prime }$ if and only if $f(x)\in f(x^{\prime })$ for all $x,x^{\prime }\in X$. When $Y$ is transitive, surjectivity gives the stronger recursive descriptionIndeed 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 the map and its transitive range are unique.

The necessary and sufficient condition is that $⟨X,\in ⟩$ be 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 $x,x^{\prime }\in X$ have distinct predecessor sets $x\cap X$, $x^{\prime }\cap X$. Sufficiency is the Mostowski collapse theorem.

For necessity, an isomorphism transfers well-foundedness from its transitive range, where Foundation applies. For extensionality, if $x\cap X=x^{\prime }\cap X$, the recursive descriptions give $f(x)=f(x^{\prime })$, and injectivity gives $x=x^{\prime }$. 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 $X$.