What's the involution elements in the permutation group of the natural number?

Denote $S_{\mathbb{N}}$ as the permutation gourp of natural numbers, i.e. all of the bijections $f:\mathbb{N}\rightarrow\mathbb{N}$. I wonder how the involution elments in $S_{\mathbb{N}}$ look like?

Actually this quetion rises when I want to figure out what's the explict formula of number seris $a_{a_n}=n$,and this leads to the kind of functions in $S_{\mathbb{N}}$ of the form $f(f(x))=x$.

Note that if $f(m)=n$, then $f(n)=m$, which states $f$ is just compsitions of a series disjoint 2-permutations, i.e. $f=(i_1i_2)(i_3i_4)\ldots (i_mi_{m+1})\ldots$,where all of the $i_k$ is distinct and this series could be infinite.Is that right? Are there any relation to the elments of order 2 in the finite permutation group $S_n$ ?

EDIT:Thanks for the existing answer, and it seems that I get some defination mixed up. It seems that I'm not looking for idempotent elements but the involution ones, i.e. the order 2 elements in $S_{\mathbb{N}}$. So sorry about that. I have corrected it above.

Solution 1:

Besides the identity, there is no idempotent permutation.

Proof: If $f$ is idempotent, then it verifies $f^2=f$.

Appy the inverse $f^{-1}$ of $f$ (which is known to exist since $f$ is a bijection) to that equation, and you get $$f=\text{Id}$$