Let $M$ a non-well founded model of Finite $\sf ZF$, which is $\sf ZF$ with axiom of infinity replaced by the axiom stating that all sets are finite. So there must be a set $\zeta$ that $M$ thinks it's a natural yet it possess's an infinite descending membership chain as seen from the outside of $M$, i.e. we have $\{ \zeta-1, \zeta-2, \zeta-3, ...\} \subset \zeta $, where each $\zeta -n = \zeta -n-1 \cup \{\zeta-n-1\} $ for all $n \in \mathbb Z$.


Since $M$ is a model of $\sf ZF$, so there must be stages: $$....,V\_{\zeta-2}, V\_{\zeta-1}, V\_\zeta, V\_{\zeta + 1}, V\_{\zeta+2},...$$


Now we know that some models of $\sf ZF$ can have external downward rank-shifting automorphisms $j$ on them, i.e. $j(V\_\alpha) =V\_\beta$ where $\beta < \alpha$ for some non-standard ordinal $\alpha$.



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> Can there be a non-well founded model $M$ of Finite $\sf ZF$, that admits an external automorphism $j$ such that: $j(V\_{\zeta+n})= V\_{\zeta +n -1}$ ?
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About external automorphism on non-well founded model of Finite ZF?
The answer is no, there cannot be even a single instance of this for an automorphism $j:M\to M$. The reason is that if $j(V\_{n+1})=V\_n$ for some (possibly nonstandard $n$), then we would have $j(n+1)=n$, and so $j$ does not respect that one of these numbers is even and the other odd. So the map couldn't be truth-preserving on $M$.


