TITLE: Is there any criteria for whether the automorphism group of G is homomorphic to G itself?
QUESTION [14 upvotes]: In the elementary group theory we know that for the symmetric groups $S_n$, except $S_6$, we have $Aut(S_n) \cong S_n$. Then the following question is natural:
What is the necessary and sufficient condition for $G$ such that $Aut(G) \cong G$?

REPLY [10 votes]: There does not exist a reasonable necessary and sufficient condition for an infinite centerless group to be complete. More precisely, letting $V$ be the set-theoretic universe, there exists an infinite complete group $G \in V$ and a $c.c.c$ notion of forcing $\mathbb{P}$ such that $G$ has an outer automorphism in the generic extension $V^{\mathbb{P}}$. An example can be found in :
S. Thomas, The automorphism tower problem II, Israel J. 
Math. 103 (1998), 93-109.
In fact, it can be shown that the group $G$ in this paper satisfies the stronger property that $G \not \cong Aut(G)$ as abstract groups in $V^{\mathbb{P}}$. In other words, there does not even exist a ''non-canonical isomorphism''.
For more on the ``nonabsoluteness'' of the height of automorphism towers, see:
J. Hamkins and S. Thomas, 
Changing the heights of automorphism towers, Annals Pure Appl. Logic 102 (2000), 139-157.