Proof check: Set of bijections from $\mathbb{N}\to \mathbb{N}$ is not countable. I've seen other answers to this question and wanted to take another approach to it. I'd appreciate some feedback:
Suppose said set was countable, then we can enumerate each bijection, say as $S:=\{f_n: \mathbb{N} \to \mathbb{N} | n\in \mathbb{N}\ \text{and } f_n \text{ is a bijection}\}$. Since each $f_n$ is bijective, can create the correspondence to its unique inverse:
$$f_n \longleftrightarrow f_n ^{-1}$$
Consider the set of bijective inverses, $S':=\{f^{-1}_{n}: \mathbb{N} \to \mathbb{N} | n\in \mathbb{N}\ \text{and } f^{-1}_{n} \text{ is the inverse of} f_n \}$. Then it must be that
$$S = S'$$,
otherwise, there is some inverse function $f_m^{-1}$ not in S, which would contradict the countability of $S$.
On the other hand, $S \subset S'$ means that for an arbitrary $n$, $f_n \in S'$, so there is some $m\in \mathbb{N}$ such that
$$f_n(x) = f_m ^{-1} (x) $$
for each x. This means that,
$$f_m(f_n (x)) =x $$
So that $f_m$ is the inverse of $f_n$; but by construction of S', it must be that $m=n$. Since $n$ was arbitrary we have that each bijection is its own inverse, which means that for each $n$, $f_n=Id_{\mathbb{N}}$, the identity function on $\mathbb{N}$. Thus, $$S = \{Id_{\mathbb{N}}\}$$ a contradiction.
 A: You can construct some bijection like this, for all pairs of consecutive numbers, either let them unchanged (coded $0$) or swap them (coded $1$).
Example :
$\begin{array}{l:ll|ll|ll|ll|ll|l}n&0&1&2&3&4&5&6&7&8&9&\cdots\\f(n)&0&1&3&2&4&5&6&7&9&8&\cdots\\\text{code}&&0&&1&&0&&0&&1&\cdots\end{array}$
You can see that $f$ is a bijection which is its own inverse since both identity and swaps over two elements are their own inverse.
This makes you last argument ($S=\{Id_N\}$) invalid.
It is also clear that any sequence of $0$ and $1$ uniquely defines such bijections, therefore there are $\#(\{0,1\}^\mathbb N)=2^{\aleph_0}$ such bijections, and consequently at least $2^{\aleph_0}$ bijections from $\mathbb N$ into itself, making it uncountable.
On the other hand there are less bijections from $\mathbb N$ into itself than simply functions from $\mathbb N\to\mathbb N$ whose cardinal is also $2^{\aleph_0}$ (see Encode each $n_1,n_2,n_3,...∈N^N$ by an inﬁnite sequence of 0s and 1s with inﬁnitely many 0s, and give a proof that $N^N$ is equinumerous with $R$.).
This time you have an injection so the number of bijection is at most $2^{\aleph_0}$, and since we proved above it is also at least this, then it is equal to $2^{\aleph_0}$.
A: The step where you are wrong is this:

So that $f_m$ is the inverse of $f_n$, but this is to say that $m=n$.

Non sequitur! If one bijection is the inverse of another, it does not follow that function and other (=inverse) function are equal.
A: Ok I'll answer here, not with the answer (I gave it in the comments under the question and your question has some duplicates on the site) but with an explanation of why your argument is wrong, which is far more instructive in my opinion.
You say: suppose the set of bijections $S$ is countable, so $S=\{ f_n \text{ bijections} \}$. Then map this set by sending $f_n$ to $g_n = f_n ^{-1}$ in $S'$. It's easy to see that $S=S'$. Now you say: given $f_n$, consider a $g_m$ such that:
$ f_n (x)= g_m (x) \text{ for every } x.$
But then you see that it must be $g_m = f_n$ BY DEFNITION. If two functions send the same elements in the same place they are equal. So by inverting you find $x=x$, which yeah it's true but meaningless. Your error is in thinking the functions in $S'$ as inverses, that's what put you off. If you wanted to use inverses, than you'd want a $f_m ^{-1}$ such that:
$$ f_m ^{-1} (x) = f_n (x) \quad \forall x \in \mathbb{N}.$$
But beware: this means that $f_m$ sends $f_n (x)$ to $x$, because the you want the inverse to do the reverse. So in the end $f_m= (f_n ^{-1})^{-1} = f_n$.
