Assume $G$ is a finite group and $\Psi \in \text{Aut(G)}$ such that for every $g \in G:\Psi(g)=g \implies g=e_G$.
Show that every element of $G$ can be written as $a^{-1}\Psi(a)$ for some $a \in G$.
I tried to use some basic results, however, these were not helpful, besides I don't know how the condition $g \in G:\Psi(g)=g \implies g=e_G$ will help us.
Any hint is appreciated.
The additional condition is required, as without it the map $\phi: G\rightarrow G$ defined by $g\mapsto\Psi(g)g^{-1}$ is not injective.
We want to prove that the map $\phi: G\rightarrow G$ is surjective. As $G$ is a finite set, this map is surjective if and only if it is injective. We shall prove that the map is injective.
To see that it is injective, suppose $\phi(g)=\phi(h)$, so $\Psi(g)g^{-1}=\Psi(h)h^{-1}$. This rearranges to give $\Psi(h^{-1}g)(h^{-1}g)^{-1}=1$, and so (by the condition) $h^{-1}g=1$, so $h=g$. Hence, the map is injective. It is therefore surjective, by the above, as required.