In a previous problem, I already proved that $\phi_{g}$ is an isomorphism, for $\phi_{g}(x)=gxg^{-1}$ so knowing that $\phi_{g} = gxg^{-1}$ is an isomorphism will certainly help here. Anyway, this is what I get:
Let $a,b \in G$ $\psi(ab) = \phi(ab)$ If $\phi_{g} = gxg^{-1},$ then $\phi_{gh} = ghx(gh)^{-1} = ghxh^{-1}g^{-1}$ $\psi(ab)=\phi_{ab}=abxb^{-1}a^{-1}$
Then I get stuck. I don't know where to go from here. I know that I need to show $\psi(ab)=\phi_{ab}=\phi_{a}\phi_{b}=\psi(a)\psi(b)$ but we don't know that G is abelian so I can't rearrange everything to my liking
If we want to prove that any two maps are same then we should show that both maps agree on all arguments where they are defined.
So to prove:
$\psi(gh)=\psi(g)°\psi(h)$
i.e, $\phi_{gh}=\phi_g ° \phi_h$
We have to show that $\phi_{gh}(x)=\phi_g ° \phi_h(x) \ \forall x \in G$
Take any $x\in G$ ,
$\phi_{gh}(x)=(gh)x(gh)^{-1}=ghxh^{-1}g^{-1} \tag{1}$
And
$\phi_g ° \phi_h(x)=\phi_g(\phi_h(x))=\phi_g(hxh^{-1})=ghxh^{-1}g^{-1} \tag{2}$
From (1) and (2), Conclusion follows .