Suppose $f, g$ are set-maps from some set $S$ to itself such that $f \circ g = g \circ f$. (1) What can we say about $f, g$? There seems to be surprisingly little.
If not, consider the following specialization: $f$ is a set-map from $S$ to itself such that for all maps $g$ (also from $S$ to itself) we have $f \circ g = g \circ f$. (2) Can we now say anything about $f$?
If still not, consider the following generalization (?): $f, g$ are set-maps from $S$ to itself, $T$ to itself respectively, such that for all maps $\varphi: S \rightarrow T$, we have $\varphi \circ f = g \circ \varphi$. (3) What can we now say about $f, g$?
(The original problem, made most obvious by (3), is to find all natural transformations of the identity functor $Set \rightarrow Set$. These questions are just what I found to be the obvious things to think about. Unfortunately, I'm having a hard time answering any of them and would appreciate some help.)
How about this: if $f(x)=x$ then $f(g(x))=g(f(x))=g(x)$. To put it in words, $g$ maps each fixed point of $f$ to a fixed point of $f$.
And then another thought: since $f^2 \circ g = f \circ g \circ f = g \circ f^2$, it follows that $g$ maps each fixed point of $f^2$ to a fixed point of $f^2$. And to translate that into a statement about $f$ itself, $g$ maps each point of period at most $2$ for $f$ to a point of period at most $2$ for $f$.
And then a further thought: by a similar argument for each natural number $n$, $g$ maps each point of period at most $n$ for $f$ to a point of period at most $n$ for $f$.
And then, by symmetry, $f$ maps each point of period at most $n$ for $g$ to a point of period at most $n$ for $g$.
Perhaps there's even more...