Endomorphism of the symmetric group of the set of positive integers via action on the prime numbers

Question

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?

Answer 1

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).

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.