For a positive integer $n$, let $p\_n$ denote the $n$-th prime number.
Further let $f: {\rm Sym}(\mathbb{N}) \rightarrow {\rm Sym}(\mathbb{N})$
be the monomorphism which maps a permutation $\sigma$ to the permutation
$f(\sigma)$ which maps any prime number $p\_n$ to $p\_{n^\sigma}$ and which is
a homomorphism of the monoid $(\mathbb{N},\cdot)$.


Denoting by $f^{(k)}$ the $k$-fold iteration of $f$, is it true that the
trajectory $(f^{(k)}(\sigma))\_{k \in \mathbb{N}}$ is cyclic only if
$\sigma$ is the identity?


Further, is it even true that
$$\bigcap\_{k=0}^\infty f^{(k)}({\rm Sym}(\mathbb{N})) \ = \ 1?$$
If the latter is false, can one explicitly describe a non-identity element
of this intersection?



Endomorphism of the symmetric group of the set of positive integers via action on the prime numbers
If $\sigma$ is the identity, then so is $f(\sigma)$.


Suppose that $\sigma$ is not the identity. Let $m\_k$ be the smallest non-fixed point of $f^{(k)}(\sigma)$. 
It is clear that $m\_0\geq 1$ is some finite integer, and for any $k\geq 0$, $m\_{k+1} = p\_{m\_k} > m\_k$.
In particular, we have $m\_k\geq q\_k$, where $q\_0=1$ and $q\_{i+1}=p\_{q\_i}$ for $i\geq 0$ ([A007097](https://oeis.org/A007097)).


It follows that the trajectory of $f^{(k)}(\sigma)$ cannot be cyclic unless $\sigma$ is the identity.


Now, for every $k\geq 0$, all elements of $f^{(k)}(\mathrm{Sym}(\mathbb N))$, except the identity, have smallest non-fixed point $\geq q\_k$. Hence, the intersection of $f^{(k)}(\mathrm{Sym}(\mathbb N))$ consists of the identity permutation only.


