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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 $\alpha$.

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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  • $\begingroup$ Since finite set theory, suitably axiomatized (i.e. with $\in$-induction, not just regularity), is bi-interpretable with PA, this question is equivalently cast as a question about automorphisms of PA. $\endgroup$
    – Joel David Hamkins
    Commented Aug 5, 2022 at 19:31

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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 $M$.

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    $\begingroup$ By the same argument, if $j:M\rightarrow M$ is an external automorphism and $\kappa,\lambda$ are $M$-ordinals with $j(\kappa)=\kappa+\lambda$ (this is equivalent to $j(V_\kappa)=V_{\kappa+\lambda}$) and $\lambda<\omega^M$ then $\lambda$ must be divisible by every standard natural number $n$. More generally, $j(\kappa)=\kappa+\lambda$ (with no size bound on $\lambda$) implies a lot of homogeneity for $\lambda$. $\endgroup$
    – Noah Schweber
    Commented Aug 5, 2022 at 18:55
  • $\begingroup$ (Whoops, I missed that this was about finite set theory; my previous comment is a bit silly in that context (where "$\omega^M$" simply isn't a thing), but does hold for nonstandard models of ZF.) $\endgroup$
    – Noah Schweber
    Commented Aug 5, 2022 at 19:33
  • $\begingroup$ But is that the only reason? I mean can we have $j(V_{n+2})=V_n$ $\endgroup$
    – Zuhair Al-Johar
    Commented Aug 5, 2022 at 19:35
  • $\begingroup$ @ZuhairAl-Johar As my comment says, no we cannot: just think mod 3 instead of mod 2. (Basically, the same reason generalizes to give the fact in my original comment.) $\endgroup$
    – Noah Schweber
    Commented Aug 5, 2022 at 19:36
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    $\begingroup$ @ZuhairAl-Johar "but this way we'll have no automorphism at all, because every time we can take mod n, and violate any automorphism." No, that's incorrect: this trick only works for standard $n$s. Models with nontrivial automorphisms do exist, but they have to move ordinals by nonstandard amounts. The point is that in order to get (any version of) the analogue of the argument in this answer to work, you need the relevant mod property to be actually definable. $\endgroup$
    – Noah Schweber
    Commented Aug 5, 2022 at 19:41

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