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.