TITLE: What is the largest set for which its set of self bijections is countable?
QUESTION [30 upvotes]: I recently came across a problem which required some knowledge about the self bijections of $\mathbb{N}$, and after looking up how to construct some different bijections I came across the result that the set of self bijections of $\mathbb{N}$ is uncountable. 
And this got me wondering, what is the largest set for which its set of self bijections is countable? This obviously holds true for any finite set, but what is the last example of a set whose set of self bijections is countable?

REPLY [2 votes]: Let's write $X!$ for the set of bijections $X\to X$.
It is true that 
$$
|X| < |Y| \implies | X! | \le | Y! |.
$$
However, it is not true that strict equality always holds. This is independent of set theory (ZFC). See below for details. 
First, though, to address the question: There is no "largest [size of] set for which its set of self bijections is countable".
For finite $X,Y$, clearly the set of bijections $X\to Y$ are a subset of $Y^X$, the set of all functions $X\to Y$; so the set of bijections is finite.
The next largest size is $\aleph_0$, the cardinality of $X = \Bbb N$. The bijections $X!$ from this set to itself have cardinal $2^{\aleph_0}$, as shown below, so already the number of bijections is uncountable. If $Y$ is any larger set, then $|X!| = 2^{|X|} \le 2^{|Y|} = |Y!|$, so the cardinality of the bijections of $Y$ is uncountable too.

It is not true that $|X| < |Y|$ implies that there are more bijections $Y\to Y$ than there are $X\to X$, for infinite $X,Y$. This is independent of ZFC, so it's not likely to be "obvious";/ We have: 
$$2^{|X|} \le (\text{# of bijections } X\to X) \le |X|^{|X|} = 2^{|X|},\tag{*}
$$
To see the first inequality, consider the injection 
$$f\mapsto \big(x\mapsto (f(x), x)\big) \colon 2^X \to (\text{bijections } X \to 2\times X).$$
Similarly, (*) holds for $Y$. 
However, in some models of ZFC, there are infinite $X,Y$ with $|X| < |Y|$ but $2^{|X|} = 2^{|Y|}$. In other models, there are no such $X,Y$ and the property is true. Assuming ZFC is consistent, neither is provable.