I was reading "Functional Equations and How to Solve Them" by Small and the following comment pops up without much justification on p. 13:
If $a(x)$ is an involution, then $f(a(x))=f(x)$ has as solutions $f(x) = T\,[x,a(x)]$, where $T$ is an arbitrary symmetric function of $u$ and $v$.
I was wondering why this was true (it works for examples I've tried, but I am not sure $(1)$ how to prove this and $(2)$ if there's anything obvious staring at me in the face here).
Any function f(x) that is a solution to your functional equation f(a(x)) = f(x) must satisfy the property that it is unchanged when you plug in a(x) instead of x. In addition, f(x) clearly must be a function f(x) = T[x, a(x)] depending on the x and a(x); the question is to see why T must be symmetric.
Now, T[x, a(x)] = f(x) = f(a(x)) = T[a(x), a(a(x))] = T[a(x), x] since a(x) is an involution. In particular, this means that T must be symmetric in its two variables.