If $f(\sigma)$ is the identity, then so is $\sigma$.

Suppose that $f(\sigma)$ is not the identity. Let $m_k$ be the smallest non-fixed point of $f^{(k)}(\sigma)$. It is clear that $m_1>1$ and for any $k\geq 1$, $m_{k+1} = p_{m_k} > m_k$.

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 greater than or equal to $q_k$, where $q_0 = 1$, $q_{i+1} = p_{q_i}$ ($i\geq 0$). Hence, the intersection of $f^{(k)}(\mathrm{Sym}(\mathbb N))$ consists of the identity permutation only.

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.