Equational identities of involutions

Question

Let $S$ be a set, and let $f$ be an involution on $S$ that is not the identity. Consider the algebra $(S;f)$. I conjecture that the equational identities of that algebra are generated by the single equation $f(f(x))=x$. Is this true, or is there a counterexample where that identity is not sufficient?

Answer 1

Given that we only have one unary operation $f$, any non-trivial identity has the shape $$f^n(x) = f^m(x)$$ or $$f^n(x) = f^m(y).$$ In the first case, if $n$ and $m$ have the same parity, then, by applying $f^2(x)=x$ to both terms, this is equivalent to $f^2(x)=x$, from which we started; otherwise, the displayed identity is equivalent to $f(x)=x$.
Hence your conjecture is correct.

In the second case, it reduces to $x=y$ (if $n$ and $m$ are even), $f(x)=f(y)$ (if $n$ and $m$ are odd), whence $x=y$ or $f(x)=y$ (if the parity of $n$ and $m$ is not the same).
In any case, it follows that $|S|=1$.