# Consistency of the Failure of Choice without Foundation

I'm working through the 2013 edition of Kunen's Set Theory, and I'm having some trouble with Exercise II.9.10, which shows the consistency of $ZF$ without Foundation and the negation of the axiom of choice relative to $ZFC$ with Foundation replaced by the assertion that there exists an infinite set of Quine atoms. Replacing Kunen's notation with what I think is more standard, he defines for any transitive set T $$V_0(T)=T \\ V_{\alpha+1}(T)=\mathcal{P}(V_\alpha) \\ V_\gamma(T) = \bigcup_{\beta<\gamma}V_\beta(T) \text{ for limit \gamma}$$

and $V(T)$ is the union of $V_\alpha(T)$ over all ordinals. Then, letting T be any infinite set of Quine atoms, he defines a subclass $M$ of $V(T)$ consisting of all $y$ such that there exists a finite $A\subseteq T$ such that for all permutations $\pi$ of $T$ which fix everything in $A$, if $\hat{\pi}$ is the extension of $\pi$ to an automorphism of $V(T)$, then $\hat{\pi}(y)=y$.

It is apparent to me why $M$ satisfies all the set existence axioms of $ZF$, why all the well-founded sets of $M$ can be well-ordered, and why $T$ is amorphous in $M$ and therefore contradicts the well-ordering theorem, but Kunen also claims that M is transitive. However, it seems to me that any permutation of T should induce a permutation of $V_1(T)$, the full power set of T, and so any automorphism of $V(T)$ should fix $V_1(T)$, so $V_1(T)\in M$, letting $A$ be empty. However, $V_1(T)$ contains many sets which are not in $M$, such as all infinite coinfinite subsets of $T$, so $M$ is not transitive and there is no reason to expect Extensionality to hold.

What am I missing here?

This might be a typo, or an attempt to avoid complex introduction of the subject matter. But $M$ should be all constructed by induction by taking at each successor step $M_{\alpha+1}$ to be those subsets of $M_\alpha$ which are fixed by all automorphisms which fix some finite set of atoms pointwise.