Calculating cardinals using Schröder–Bernstein theorem

Question

I'm trying to calculate the cardinality of these two sets. I think the answer is ℵ, and I want to prove it by finding proper injective functions, and using Schröder–Bernstein theorem. Any help will be much appreciated.

F = {f ∈ N → N | f is a bijection}

G = {f ∈ N → N|f ◦ f = idN}

Answer 1

$F$ is the set of permutations of $N$ and $|F| = |\mathscr P(N)| = c $ the cardinality of the reals. $c \ne \aleph_0$ and if you accept the continuum hypothesis then $c = \aleph_1 = 2^{\aleph_0}$. https://en.wikipedia.org/wiki/Aleph_number

$G$ is a subset of permutations (i,e. $G \subset F$) corresponding to swapping elements of $N$. Proof....

Let $g(a) = b $ Then $ a = g(g(a)) = g(b) $, i.e. $g(a) = b \implies g(b) = a$.

So, $|G| \le |F|$ (and there is an injection from $G \to F$)

This is a bit long for a comment, but is only a partial answer as I don't see how to do an injection the other way (but I think you should be able to prove directly that $|G| = |\mathscr P(N)| $)

Answer 2

SKETCH: $G\subseteq F$, so the inclusion map gives you one of the injections needed to apply the Cantor-Schröder-Bernstein theorem. Now we need an injection $\varphi:F\to G$ that assigns to each permutation of $\Bbb N$ a unique involution on $\Bbb N$. An involution is a product of disjoint transpositions, so we need to figure out a way to code an arbitrary permutation as such a product.

The idea is to split $\Bbb N$ into $E=\{2n:n\in\Bbb N\}$ and $O=\{2n+1:n\in\Bbb N\}$, the sets of even and odd natural numbers, respectively. (Note that $0\in\Bbb N$ here.) Let $f\in F$ be any permutation of $\Bbb N$. We’ll build $\varphi(f)$ as a product of transpositions, each transposing a member of $E$ and a member of $O$ and each encoding one value of $f$. Specifically, for each $n\in\Bbb N$ let $\tau_n^f$ be the transposition that interchanges $2n$ and $2f(n)+1$; the transpositions $\tau_n^f$ for $n\in\Bbb N$ are pairwise disjoint, so their product is an involution. What, exactly, is their product? It’s the function $g:\Bbb N\to\Bbb N$ defined as follows:

$$g(n)=\begin{cases} 2f\left(\frac{n}2\right)+1,&\text{if }n\in E\\ 2f^{-1}\left(\frac{n-1}2\right),&\text{if }n\in O\;. \end{cases}\tag{1}$$

Thus, we want to set $\varphi(f)=g$. If you think of $f$ as a set of ordered pairs, each $\langle n,m\rangle\in f$ is encoded in $g$ as a transposition of $2n$ and $2m+1$,, and splitting $\Bbb N$ into the two infinite sets $E$ and $O$ ensures that we have enough ‘room’ to encode each value of $f$ independently of the encodings of the other values.

I’ll leave it to you to go through the details of verifying that $(1)$ really does work as claimed.