Does the functor $\mathrm{Aut}$ have an adjoint I am studying for a group theory exam, trying to present the subject categorically.
In order to study for this I am trying, to in every exercise spot if there are any functors involved and trying to find if they have adjoints. One such functor I have stumbled upon which has stumped me is
$$\mathrm{Aut} \colon \quad G \mapsto \mathrm{Aut}(G), \quad (f \colon G \rightarrow H) \mapsto \mathrm{Aut}(f)(\phi)=f\circ\phi\circ f^{-1} \,.$$
From the category $\mathbf{Grp}_{\mathrm{iso}}$ to  itself.
Intuitively I would want to say that studying the automorphism groups should be weakly equivalent to studying the underlying group, is there any adjunction to formalise this hunch? And if not, can we prove why?
Also is there any way to study $\mathrm{Aut}$ as a functor, without restricting to isomorphism, because this makes the category a lot less interesting?

In the category $\bf{Grp}_{\text{iso}}$ with groups as objects and group isomorphisms as morphisms, there are no "interesting" morphisms. Indeed, $\operatorname{Hom}(A,B)$ is non-empty only if $A$ and $B$ are isomorphic.
For an adjoint functor $F$ to $\operatorname{Aut}\colon \bf{Grp}_{\text{iso}}\to \bf{Grp}_{\text{iso}}$, we demand either
$$ \operatorname{Hom}(A,\operatorname{Aut}(B))\cong \operatorname{Hom}(F (A),B)$$
or
$$ \operatorname{Hom}(\operatorname{Aut}(A),B)\cong \operatorname{Hom}(A,F(B)).$$
In particular both sides are empty or both are non-empty.
Thus we want $$A\cong \operatorname{Aut}(B)\iff F(A)\cong B.$$
To show that this is impossible, it suffices to find two groups $B_1,B_2$ with $B_1\not\cong B_2$ and $\operatorname{Aut}(B_1)\cong \operatorname{Aut}(B_2)$. That's easy: For example, the trivial group and the group of order $2$ both have trivial automorphism group. Or $\Bbb Z$ and the group of order $3$ both have the group of order $2$  as automorphism group
