On Kunen's Set Theory, Exercise II.9.3 The exercise is as follows:

Suppose that $F : \mathbf{V} \to \mathbf{V}$ is a permutation of $V$ and $E = E_F$. Define $i : \mathbf{V} \to \mathbf{V}$ by recursion on $\in$ so that $i(x)$ is the $y$ such that $F(y) = \{i(w) : w \in x\}$. Prove that $i$ is $1$-$1$, and is an isomorphism between $(\mathbf{V},\in)$ and $(\mathbf{WF}^{(V,E)},E)$.
Also, $\mathsf{AC} \leftrightarrow \mathsf{AC}^{(\mathbf{V},E)}$.
Also, the model $(\mathbf{V},E)$ satisfies the axiom that every relation $R$ on a set $A$ is isomorphic to a relation on a well-founded set.

Note that here, $x \, E_F \, y$ iff $x \in F(y)$.
I'm unable to comprehend how does $i$ work or "look like". Intuitively, I try to think of $i$ as an embedding (as hinted in the previous paragraph), so $i(x)$ should be a set that "contains" (in the sense of $E$) all $i(w)$'s, where $w \in x$. But this does not help me in proving the claims in the exercise.
Thanks in advance.
 A: The idea is that for any set in $(V,\in)$ we can construct an isomorphic copy of it in $(V,E)$ using $\in$-recursion. The isomorphic copy of $x$ is $i(x).$ First we construct the empty set, which is the set with no $E$-elements. This is just $F^{-1}(\emptyset),$ so $i(\emptyset) = F^{-1}(\emptyset).$ Then, if we have constructed all the elements of a given set and want to construct the set itself, we just take $F^{-1}(\{i(w): w\in x\}),$ which is the set whose $E-$elements are the isomorphic copies of all the elements of $x.$ So the first part is just to prove that this works as advertised, and that furthermore every well-founded set in $(V,E)$ is in the range of $i$ (think about the Mostowski collapse of the relation $E$ on the given set).
As for the other parts, it is good to think about what a relation in $(V,E)$ looks like. A relation $R$ on $X$ should be a set whose elements are ordered pairs of elements of $X,$ but what are ordered pairs? Write $$ (x,y)^E = F^{-1}\{F^{-1}(\{x\}), F^{-1}(\{x,y\})\},$$ i.e. (according to $(V,E)$) the set whose elements are the set whose only element is $x$ and the set whose elements are $x$ and $y$. So $R$ is a relation on $X$ means that $F(R)$ is a set consisting of elements of the form $(x,y)^E$ where $x,y\in F(X).$
For any relation $R$ on $X$ (according to $(V,E)$), we can find a corresponding relation (according to $(V,\in)$) on $F(X)$ $$ \tilde R = \{(x,y): (x,y)^E \in F(R)\},$$ and we know from the first part that $(V,E)$ has an isomorphic copy of this, $i(\tilde R),$ which according to $(V,E)$ will be a relation on a well-founded set that is isomorphic to $R.$
