Let $A$ be the group of anti-automorphisms, $B=Acup operatorname{Aut}(G)$, prove $Bcong Atimes mathbb{Z}/2mathbb{Z}$

Let $A=\{\phi\in S(G):\forall_{a,b\in G}: \phi(ab)=\phi(b)\phi(a)\}$ be the set of anti-automorphisms of a non-abelian group $G$, define

$B=A\cup \operatorname{Aut}(G)$, with $\operatorname{Aut}(G)$ being
the group of automorphisms of $G$.

Prove that $B\cong \operatorname{Aut}(G)\times$
$\mathbb{Z}/2\mathbb{Z}$

During my proof that $B$ is a group, I discovered some properties that make it logical that $\mathbb{Z}/2\mathbb{Z}$ is included here. I also thought that $|A|=|\operatorname{Aut}(G)|$ since you can take the same bijections, but apply the necessary conditions. What I noted:

1. $A$ and $\operatorname{Aut}(G)$ are disjoint


2. $\phi,\psi\in A \implies \phi\circ\psi \in\operatorname{Aut}(G)$


3. $\phi\in A, \psi\in\operatorname{Aut}(G)\implies \phi\circ\psi\in A$ and $\psi\circ\phi\in A$

This gives us that if you compose an even number of elements from $A$, you will end up with an element from $\operatorname{Aut}(G)$, and if you compose an uneven number of elements from $A$, you will end up with another element in $A$. This gives me an intuitive feeling that $\mathbb{Z}/2\mathbb{Z}$ is involved, but I can't quite find the right bijection to completely prove the statement.

Is this the way to go about it, or should is there a different approach?