TITLE: Nonabelian finite $G$ such that $O(g) \subseteq Z(G)g, \forall g \in G$
QUESTION [8 upvotes]: Is there any nonabelian finite group $G$ such that:
$$O(g) \subseteq Z(G)g, \forall g \in G \tag 1$$
where $O(g)$ is the orbit "by $g$" of the natural action of $\operatorname{Aut}(G)$ on $G$, namely $\sigma \cdot g := \sigma(g)$?

REPLY [2 votes]: There are three (isomorphism classes of) groups of order 64 with this property, which are $\mathtt{SmallGroup}(64,i)$ with $i=68,69,116$ in the small groups database.
They all have commutator subgroup $[G,G] \cong C_2 \times C_2$ with $G/[G,G] \cong C_2 \times C_2 \times C_4$. The first two have $Z(G) \cong C_2^3$ and automorphism groups of order $512$, and the third has $Z(G) \cong C_2 \times C_4$ and automorphism group of order $128$.