Let $G$ be a group and $G^{op}$ denotes its opposite group. It is well-known that the functor $F$ from $Grp$ to itself, defined by $$ \begin{aligned} G&\mapsto G^{op}\\ x&\mapsto x^{-1}\\ \phi &\mapsto [x \mapsto \phi(x)^{-1}], \end{aligned} $$ where $\phi(x)^{-1}$ denotes the inverse element of $\phi(x)$ in the image of $\phi$, and not the inverse map of $\phi$, is an involution of the category.
My question is, are all anti-homomorphisms of from a group $G$ to itself of the form $F(\phi)$, for a homomorphism $\phi:G\rightarrow G$?
Let's separate those maps, and connect them in a more proper way.
An anti-homomorphism $A\overset{\text{anti}}\longrightarrow B$ between (semi)groups is just an ordinary homomorphism $A\to B^{op}$ where $B^{op}$ has the same elements as $B$ but the operation is reversed.
Alternatively, we can equally well say that it's a homomorphism $A^{op}\to B$.
For every group $G$, we have the anti-homomorphism $i_G:G\to G^{op}$ or $G^{op}\to G,\ \ x\mapsto x^{-1}$, which is actually involutive: it is basically its own inverse.
This induces a mapping $\hom(A,G)\to\hom(A,G^{op}),\ \ f\mapsto\, i_G\circ f$, and the same way it also induces its inverse: $$\hom(A,G^{op})\to\hom(A,G),\ \ f\mapsto i_G\circ f$$