# $f(a(x))=f(x)$ - functional equation

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).

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Write down what f(a(x)) is in terms of T. –  Qiaochu Yuan Jul 30 '10 at 2:42
Oh wow. This is pretty silly. Thanks- –  A B Jul 30 '10 at 2:51
The same applies if you have a map $a$ satisfying $a^n = Id$ and there is a symmetric function in $x, a(x), ... , a^{n-1}(x)$. (Note that powers here refer to iterated composition.) –  Akhil Mathew Jul 30 '10 at 2:52