Inverse of an automorphism I am being more stupid than usual: embarrassingly so. Let $G$ be a group. Let $\phi:G \rightarrow G$ be an automorphism with inverse $\psi$. Is it true (and why is it true) that $\psi(x)=\phi(x^{-1})$. 
Right now, I got myself in a gumption trap and have not thought my way out of it.
 A: This is not true in general. For instance consider the case of the identity $I_{G}:G \rightarrow G $. It is its own inverse, nevertheless it isn't true that $I_{G}(x) = I_{G}(x^{-1})$ because that would imply that $x = x^{-1}$ for every $x \in G$.
Another counterexample along the same lines. If $G$ is an abelian group then you have available the automorphism $\phi : G \rightarrow G$ given by $\phi (x) = x^{-1}$. Then in this case also $\phi$ is its own inverse and the same argument as before applies.
A: After brushing off some cobwebs, I thought I might see for which groups the identity can hold.
Claim: Let $G$ be a finite group for which there exists $\phi \in \mathrm{Aut}(G)$ for which $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$.  Then $G$ is an elementary abelian 2-group.
Proof:  If $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$, then $\phi(\phi(x))=\phi^{-1}(\phi^{-1}(x^{-1}))$ and so on.  Hence $x=x^{-1}$ and $x^2=\mathrm{id}$ for all $x \in G$, and so $G$ must be an elementary abelian 2-group. $\square$
In fact, a converse of the above is true, since each elementary abelian 2-group admits a the identity automorphism $\phi$, which satisfies the identity $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$.  But, as we will see, not all automorphisms of elementary abelian 2-groups satisfy the identity.
Claim:  Let $G$ be a finite group for which each $\phi \in \mathrm{Aut}(G)$ satisfies $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$.  Then $G \cong \mathbb{Z}_2$ or $G$ is the trivial group.
Proof:  By the previous claim, $G$ must be an elementary abelian 2-group.  Applying $\phi$ to both sides of the identity $\phi(x)=\phi^{-1}(x^{-1})$, we find that $\phi(\phi(x))=x^{-1}=x$ for all $x \in G$.  Hence $\phi^2$ is the identity automorphism for all $\phi \in \mathrm{Aut}(G)$.  Therefore, every automorphism of $G$ must not contain a 3-cycle (in fact t-cycle where $t \geq 3$).
The elementary abelian 2-group of order 4 admits an automorphism that contains a 3-cycle.  Consequently, we can prove by induction that elementary abelian 2-group of order $2^a$ for all $a \geq 2$ each admit an automorphism that contains a 3-cycle.
The claim is true for the trivial group and $\mathbb{Z}_2$ by inspection. $\square$
A: $\phi(x^{-1})=\phi(x)^{-1}$
But $\phi(x)^{-1}= \phi^{-1}(x)$, not is necesarily true .
