Bijection from $\mathbb{N} \to \mathbb{N}$ I was thinking about proofs that there are an infinite amount of bijections from $\mathbb{N} \to \mathbb{N}$, but could not think of anything else than permutations (which are already enough).
Do you have more examples of that?
 A: There are uncountably many bijections from $\mathbb{N}$ to $\mathbb{N}$. Indeed, consider the set of bijections that for every pair $2x,2x+1$, either fix both points or transpose the points. Since there are infinitely many such pairs, this gives us $2^{\aleph_0}$ bijections. There is also an easy upper bound of $\aleph_0^{\aleph_0}$, so we can conclude that the number of bijections from $\mathbb{N}$ to $\mathbb{N}$ is $2^{\aleph_0}$. I gave an example of $2^{\aleph_0}$ of them, but this is only a small sample; most of them probably cannot be described explicitly.
A: If by permutation you mean that it moves a finite number of naturals you can consider the following "cheap" trick. Consider the numbers $\bmod n$ and if a number is of the form $kn+d$ send it to $kn+d+1$ unless the number is $kn+n-1$ in which case we send it to $kn$. In this way we permute every number.
Another infinite family can be obtained by taking a partition of the naturals into an infinite number of finite sets and rotating the numbers in each partition , sending the number $n$ to the next largest number in the set, unless $n$ is the largest number in its set, then you send it to the smallest.
Another way is to select a pair of integers, switching them, and fixing the rest.
I don't know if any of these are examples of what you want.
A: For each real number in $(0,1)$ one can construct bijection as follows (I will first give an example).
Suppose the real number is 0.341. . . .
As the first decimal is 3, we use it to cyclically permute: $1\to2\to3\to1$, the second decimal 4 is used for a cyclic permutation of next 4 symbols: $4\to5\to6\to7\to4$, the third decimal 1 means we get the next number 8 as fixed point.
That is a decimal expansion such as $0.abc\ldots$ is interpreted as  a cycle of length $a$, followed by a cycle of length $b$, then a cycle of length $c$ operating on contiguous blocks of respective length from the set of natural numbers.
Remark: All these are permutations of finite order, orders dividing $10!$.
EDIT: The above has a minor snag; it gives rise to a well-defined bijection only for a real number whose decimal expansion has no $0$ as a digit. As such real numbers are still uncountable even with this defect this answers the question.  
Alternatively one can essentially use the same idea with a simple modification:
$0.abc\ldots$ should be interpreted as a a permutation that cyclically permutes
$1$ to $a+1$, then $a+2$ to $a+2+b+1$ etc. This way a zero will lead to a fixed point, and a $1$ will lead to a transposition. My final remark has a consequential change: all these permutations will be of order $11!$ or a factor of $11!$.
