Let $G$ be a group. Is $G$ abelian if $f : G \to \mathrm{Aut}(G)$, defined by $f(g) = f_g:x \mapsto g^{-1}xg$, is a homomorphism ?
I am aware of this question, but notice that I have $g^{-1}xg$ instead of $gxg^{-1}$. My condition $f_{ab} = f_a \circ f_b$ is equivalent to : $$\forall a,b \in G, \;\forall x \in G \quad b^{-1}a^{-1}xab=a^{-1}b^{-1}xba$$ Notice that $\forall x \in G, \; c^{-1}xc = d^{-1}xd \quad\implies\quad c=d$ iff $Z(G) = \{1\}$.
I tried to work with the possible counter-example $G=D_8$. I'm not sure if there is a counter-example. At least, the converse is trivial, i.e. if $G$ is abelian then $f$ is a homomorphism.
Let $\varphi_g\colon x\mapsto gxg^{-1}$. Then we know that $\varphi$ defines a homomorphism $G\to\operatorname{Aut}(G)$, with image the inner automorphisms. Also the map $f$ has the same set as image. Since $f_g=\varphi_{g^{-1}}$, the statement $f_{gh}=f_g\circ f_h$ becomes $$ \varphi_{(gh)^{-1}}=\varphi_{g^{-1}}\circ\varphi_{h^{-1}} $$ or as well, since $g$ and $h$ are arbitrary, $$ \varphi_{hg}=\varphi_g\circ\varphi_h $$ This amounts to saying that $\operatorname{Inn}(G)$ is abelian. Since $\operatorname{Inn}(G)\cong G/Z(G)$, we have that the condition is equivalent to $G/Z(G)$ being abelian, that is, $G$ nilpotent with nilpotency class $\le2$.