Part 1 asks for the case where the conjugation action $G \to \mathrm{Aut}(G)$ is injective but not surjective. So, now here is the other part (where the map is surjective but not injective).
While Part 1 is about non-cohopfian groups, this part is about non-hopfian groups.
Question: Do there exist (necessarily infinite) groups $G$ for which $\mathrm{Aut}(G) \cong G$ satisfying the following property?
Every automorphism of $G$ is inner, but the center of $G$ is nontrivial (so the conjugation action $G \to \mathrm{Aut}(G)$ is surjective but not injective).
Clearly, if $G$ satisfies this property, then it cannot be hopfian. In particular, $G$ must be infinite.
If $H$ is an infinite complete group with no element of order $2$ and also no nontrivial homomorphism to $\mathbb{Z}/2$, then $H \times \mathbb{Z}/2$ does satisfy the above property, but it is clearly not isomorphic to its own automorphism group (because it has an element of order $2$, while its automorphism group is isomorphic to $H$, which has no element of order $2$).
So, there must be another way to construct examples for the above property.
Note: The above has been moved from the original question per user1729's comment there.