Let $\omega^\omega$ be the set of finite sequences of positive natural numbers and let $\lvert x\rvert$ be the ordinal of $x\in\omega^\omega$ given by using the terms as coefficents in Cantor normal form.
Let $f:\omega^\omega\to\omega^\omega$ be a surjection that truncates sequences, i.e. sends $\forall n:\omega^n\setminus\omega^{n-1}\to\omega^{n-1}$. Then every sequence $f^n(x)$ under composition is eventually constant.
Clearly $\Bbb N$ has a monoidal action given by $F^{-n}(\emptyset):n\in\Bbb N$ over some set $X\to X$ which contains infinitely nested sequences of inclusions drawn from $\omega^\omega$, and which satisfies $\lvert F^{-1}(x)\rvert\geq\lvert x\rvert$.
It is my tentative belief that $F:\omega^n\mapsto\omega^{n-1}$ and therefore $F:\omega^\omega\to\omega^\omega$.
So there is an $f$-compatible monoid action, but it is my belief that there is also an $f$-compatible group action of $\Bbb Z$ given by $F^n(x):n\in\Bbb Z$ over sets $X$ which contain $\omega^\omega$ within their transitive closure.
Is this correct, and if so, on what set does it act? I'm trying to do here is to better picture this group action.
I believe the fact that there exists $\emptyset\in\omega^\omega$ such that $f(\emptyset)=\emptyset$ means the only such group action of $\Bbb Z$ is an identity function on the trivial group.
Moreover, I think the only set on which $F$ can be the identity function, is $\mathcal P(\omega^\omega)$. Is this correct?
If I elaborate slightly - let $F:\mathcal P(\omega^\omega)\to\mathcal P(\omega^\omega)$ be given by:
$F^{-1}(x)=\{y:f(y)\in x\}$
and consider the group action of $z\in\Bbb Z$ on $F^z$ where $z$ indicates function composition.
On what set is $F$ the identity function?
i.e. what set $X$ satisfies $F(X)=X$?