TITLE: Finding groups $G$ such that $G \cong \mathscr{S}(G)$
QUESTION [15 upvotes]: Let $G$ be a group and let $\mathscr{S}(G)$ denote the group of Inner-Automorphisms of $G$. 

The only isomorphism theorem I know, that connects a group to its inner-automorphism is:  $$G/Z(G) \cong \mathscr{S}(G)$$ where $Z(G)$ is the center of the group. Now, if $Z(G) =\{e\}$ then one can see that $G \cong \mathscr{S}(G)$. What about the converse?

That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)=\{e\}$? In other word's I need to know whether there are groups with non-trivial center which are isomorphic to their group of Inner-Automorphisms. Are there any type of classification for such type of groups?

REPLY [7 votes]: @mac - yet with this example you can create an infinite family of groups G satisfying $G \cong G/Z(G)$ by taking the direct product with any group having trivial center. Since you cannot possibly classify all groups having a trivial center, it is hopeless to expect a classification of the class of groups proposed by Chandru.