Cardinality of the set of bijective functions on $\mathbb{N}$?

The same. It suffices to show that there are $2^\omega=\mathfrak c=|\Bbb R|$ bijections from $\Bbb N$ to $\Bbb N$. Let $P$ be the set of pairs $\{2n,2n+1\}$ for $n\in\Bbb N$. (My $\Bbb N$ includes $0$.) For each $S\subseteq P$ define

$$f_S:\Bbb N\to\Bbb N:k\mapsto\begin{cases} k+1,&\text{if }k\in p\text{ for some }p\in S\text{ and }k\text{ is even}\\ k-1,&\text{if }k\in p\text{ for some }p\in S\text{ and }k\text{ is odd}\\ k,&\text{if }k\notin\bigcup S\;; \end{cases}$$

the function $f_S$ simply interchanges the members of each pair $p\in S$. Clearly $|P|=|\Bbb N|=\omega$, so $P$ has $2^\omega$ subsets $S$, each defining a distinct bijection $f_S$ from $\Bbb N$ to $\Bbb N$. Thus, there are at least $2^\omega$ such bijections. And each function of any kind from $\Bbb N$ to $\Bbb N$ is a subset of $\Bbb N\times\Bbb N$, so there are at most $2^\omega$ functions altogether. Thus, there are exactly $2^\omega$ bijections.

Both have cardinality $2^{\aleph_0}$. Note that the set of the bijective functions is a subset of the surjective functions.

To see that there are $2^{\aleph_0}$ bijections, take any partition of $\Bbb N$ into two infinite sets, and just switch between them. It is not hard to show that there are $2^{\aleph_0}$ partitions like that, and so we are done.

In addition to Asaf's answer, one can use the following direct argument for surjective functions:

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Consider any mapping $f: \Bbb N \to \Bbb N$ such that:

$$\forall n \in \Bbb N: f(2n) = n$$

Then $f$ is surjective, but for any $g: \Bbb N \to \Bbb N$ we may define $f(2n+1) = g(n)$, effectively showing that there are at least $2^{\aleph_0}$ surjective functions -- we've demonstrated one for every arbitrary function $g: \Bbb N \to \Bbb N$.