Does existence of homomorphism imply there is $\psi$ with $\psi(ab) = \psi(b) \circ \psi(a)$

Question

Let $G$ be a group and $A(G)$ its automorphism group. Suppose there is a homomorphism $\phi: G \to A(G)$. That is, for all $a, b \in G$, we have $\phi(ab) = \phi(a) \circ \phi(b)$. Must there exist $\psi: G \to A(G)$ such that, for all $a, b \in G$, $\psi(ab) = \psi(b) \circ \psi(a)$?

Answer 1

Hint:

$g \mapsto g^{-1}$ is an isomorphism $G \cong G^\text{op}$ for any group $G$.

Here $G^\text{op}$ has the same underlying set as $G$ but with multiplication done backwards: $g \cdot^\text{op} h = h \cdot g$.

In this language, $\psi$ should be a homomorphism $G \to A(G)^\text{op}$ (I'll leave it to you to check this). Can you use the isomorphism above to build such a $\psi$ if you are a given a homomorphism $\phi : G \to A(G)$?

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I hope this helps ^_^