Indeed, Cantor's argument is readily adapted. Let $\pi_1,\pi_2,\pi_3, \ldots $ be any countable sequence of permutations of $\mathbb N$ ; let us show that this sequence
does not exhaust all permutations, by constructing a permutation $\pi$ different from all
the $\pi_i$. We first define $\pi$ on the even integers inductively, then define
$\pi$ on the odd integers.

Let $X$ be any subset of $\mathbb N$ such that both $X$ and ${\mathbb N} \setminus X$
are infinite (e.g. the even integers, the prime numbers ...).

Set $\pi(0)$ to be an integer in $X$ different from $\pi_1(0)$. Set $\pi(2)$ to be an integer in $X$ not in $\lbrace \pi(0),\pi_2(2)\rbrace$. Set $\pi(4)$ to be an integer in $X$ not in $\lbrace \pi(0),\pi(2)\pi_3(4)\rbrace$. Continuing like this inductively, we define an injection from the even integers to $X$,
such that $\pi(2k-2) \neq \pi_k(2k-2)$ for any $k$ (and hence $\pi \neq \pi_k$).

Finally, the set A=${\mathbb N} \setminus \pi(2\mathbb N)$ is countably infinite ; setting
$\pi(2k-1)=$ the $k$-th element of $A$ finishes the proof.