Given a group $G$, it is shown here that:
$$Z(\operatorname{Aut}(G)) \cap \operatorname{Inn}(G) \cong H/Z(G) \tag 1$$
where $H=\lbrace a \in G \mid \sigma(a) \in Z(G)a, \forall \sigma \in \operatorname{Aut}(G) \rbrace$.
As a corollary, if $G$ is such that:
$$\sigma(a) \in Z(G)a, \forall a \in G, \forall \sigma \in \operatorname{Aut}(G) \tag 2$$
we'd have $H=G$ and then, for $(1)$:
$$Z(\operatorname{Aut}(G)) \cap \operatorname{Inn}(G) \cong G/Z(G) \cong \operatorname{Inn}(G) \tag 3$$
If we further assume $\operatorname{Inn}(G)$ finite, then (see the answer here) we'd have $\operatorname{Inn}(G) \le Z(\operatorname{Aut}(G))$: each inner automorphism of $G$ would commute with all the automorphisms of $G$.
I'm wondering whether $G$'s exist, fulfilling $(2)$ and with $\operatorname{Inn}(G)$ finite, for which then this latter fact would hold. The only I can notice is that $(2)$ is equivalent to stating that:
$$O(a) \subseteq Z(G)a, \forall a \in G \tag 4$$
$O(a)$ being the orbit by $a$ of the natural action of $\operatorname{Aut}(G)$ on $G$.
How are these $G$'s made, if there are any?
PS. Not sure if finite-groups tag is applicable.