Let a cyclic group $A=\langle a\rangle$ act on a group $G$ whose order is odd. If $[G,a^2]=1$, then $$\{x\in G: x=x^{-a}\}=\{[x,a]:x\in G\}.$$
My attempt: Since $[G,a^2]=1$, so $x^{a^2}=x$ for all $x\in G$. Therefore, $$[x,a]^a=(x^{-1}x^{a})^a=x^{-a}x^{a^2}=x^{-a}x=[x,a]^{-1}$$ for all $x\in G$. Thus, $$\{x\in G: x=x^{-a}\}\supseteq \{[x,a]:x\in G\}.$$
How do I show the other inclusion? I know that the cardinality of $\{[x,a]:x\in G\}$ is equal to $[G:C_G(a)]$, so if I can show the same for $\{x\in G: x=x^{-a}\}$, then I would be done. However, I'm not sure how to count the elements of this set.