Finding groups $G$ such that $G \cong \mathscr{S}(G)$ 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? 
 A: I can help a little, maybe not if you already thought of it : there is no point in trying to find counter examples with finite groups ; for if $G \cong \mathscr S(G)$, since $\mathscr S(G) \cong G/Z(G)$, we have $G \cong G/Z(G)$, which means those two are in bijection. Since this means $|G| = |G|/|Z(G)|$, we have $|Z(G)| = 1$ and $Z(G) = \{ e \}$.
Now while looking at infinite groups it may be interesting that you are basically trying to either find a group $G$ with a non-trivial center such that $G \cong G/Z(G)$, or trying to prove that every group with a non-trivial center is such that $G \ncong G/Z(G)$. My first approach would be to try some nasty groups as counter-examples rather than trying proofs of the second choice... 
A: @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.
A: Interesting question! It seems that there is such a group.
I have no idea about a classification of these groups, myself.
