I just learnt about the second Fraenkel model and I wanted to see how some arguments and proofs regarding it works.
So within the second Fraenkel model, let $f:A\rightarrow A$. Show that there is an $m \in \omega$ such that for all $n \geq \omega$, either $f[P_n] = P_n$ or $f[P_n] \subseteq \bigcup_{i\leq m} P_i$. Here $P_i$'s are the disjoint pairs of $A$ as defined in Jech's axiom of choice.
I have seen some solutions regarding this question, but I wish to make use of the following proposition to solve this:
"If $f: A\rightarrow A$, then there is a finite E $\subseteq A$ such that $f$ is either constant on $A-E$, or $f(a) = a,\ \forall a\in A$"
Cheers !
Your proposition cannot possibly be true, since we can consider the function $f$ such that if $P_n=\{a,b\}$, then $f(a)=b$ and $f(b)=a$. Namely, rotate each pair. This function has no fixed points and it is not constant either.
But it is true that modulo the pairs, the functions are finitary in some sense. To see this you can either work in the setting of the model itself, or prove something better and more general.
Suppose that $A$ is a Dedekind-finite set which is the countable union of pairs, $P_n$. Let $p(a)=n$ if $a\in P_n$. If $f\colon A\to A$ is a function, then $p(f(a))=p(f(b))$ whenever $f(a)=f(b)$, up to a finite mistake.
Otherwise, there are infinitely many pairs such that $f$ "separates" the members of the pair. And so we can choose the one which was sent to the lower indexed pair. So that's impossible.