Is symmetric group on natural numbers countable? I guess it is too difficult a question to ask about the cardinality of $S_{\mathbb{N}}$ so I would like to ask whether it is countable or not.

I tried to prove it is uncountable somewhat mimicking the Cantor's diagonal argument but failed.
Here's a very silly argument to show $|S_\mathbb{N}| \geq 2^\mathbb{N}$.

A fact from calculus tells us that a non-absolutely convergent series whose terms converge to zero can be reordered to take the value of any real. So, for each real $\alpha$, there is a permutation such that 
$$
\sum\frac{(-1)^{n_i}}{n_i}= \alpha
$$

So there must be at least as many permutations as reals!
Here is a slightly analytic approach, that I think isn't too similar to what anyone else has written.
Since $\mathbb Q$ is countably infinite, it suffices to show that there are uncountably many bijections $\mathbb Q \to \mathbb Q$. We exhibit an injection $\mathbb R_{&gt;0} \to S_{\mathbb Q}$.
Given a positive real $\alpha &gt; 0$, define
\begin{align*}
 f_\alpha: \mathbb Q &amp;\to \mathbb Q \\
 q &amp;\mapsto
 \begin{cases}
  q &amp; \text{if $\lvert q \rvert \le \alpha$} \\
  -q &amp; \text{if $\lvert q \rvert &gt; \alpha$}
 \end{cases}
\end{align*}
Then $f_\alpha$ is an involution, so certainly a bijection.
Moreover, it is clear that $\sup \{q \in \mathbb Q \mid f_\alpha(q) = q\} = \alpha$. This shows that $\alpha \mapsto f_\alpha$ is an injection. So indeed $S_{\mathbb Q}$ is uncountable.
I personally like this proof a lot. I find it much more intuitive when things end up being uncountable &quot;because of $\mathbb Q$&quot;, and particularly when it's because of a Dedekind-cut-like density-related construction like this, than when it's because of a diagonal argument.
(fun fact: the set of all bijections in $S_{\mathbb N}$ having no fixed points is also uncountable!)