Consider the following map $\Phi$ from bijections of the natural numbers to subsets of the natural numbers: to each bijection $\sigma$, associate its fixed point set $S(\sigma) = \{x \in \mathbb{N} \ | \ \sigma(x) = x \}$.

It is easy to see that this map is almost surjective: the only subsets which are not in the image are those whose complement consists of a single element. Indeed, a subset of the natural numbers admits a fixed point free permutation iff it does not consists of a single element. In particular, the complement of the image of $\Phi$ is countable. Since the codomain of $\Phi$ is uncountable, so therefore is the image of $\Phi$.

This shows that the set of bijections is uncountable. By the usual Schroder-Bernstein argument, it follows that it has continuum cardinality.