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 $Aut(G) \cong G$?



Is there any criteria for whether the automorphism group of G is homomorphic to G itself?
This answer is essentially a series of remarks, but ones which I hope will be helpful to you.


(1) There are two ways to interpret the condition that $G$ be isomorphic to its automorphism group: canonically and non-canonically.


a) Say that $G$ is [complete](http://en.wikipedia.org/wiki/Complete_group) if every automorphism of $G$ is inner (i.e., conjugation by some element of $G$) and $G$ has trivial center. In this case, there is a canonical isomorphism $G \stackrel{\sim}{\rightarrow} \operatorname{Aut}(G)$. 


The linked wikipedia article gives some interesting information about complete groups. As above, by definition having trivial center is a necessary condition; all nonabelian simple groups satisfy this. On the other hand, an interesting sufficient condition is that for any nonabelian simple group $G$, *its* automorphism group $\operatorname{Aut}(G)$ is complete, i.e., we have canonically $\operatorname{Aut}(G) = \operatorname{Aut}(\operatorname{Aut}(G))$.


b) It is possible for a group to have nontrivial center and outer automorphisms and for these two defects to "cancel each other out" and make $G$ noncanonically isomorphic to $\operatorname{Aut}(G)$. This happens for instance with the dihedral group of order $8$.


2) It seems extremely unlikely to me that there is a reasonable necessary and sufficient condition for a general finite group to be isomorphic to its automorphism group in either of the two sense above. 


But a lot of specific examples are certainly known: see for instance 


<http://en.wikipedia.org/wiki/List_of_finite_simple_groups>


in which the order of the outer automorphism group of each of the finite simple groups is given. So, for instance, exactly $14$ of the $26$ sporadic simple groups have trivial outer automorphism group, hence satisfy $G \cong \operatorname{Aut}(G)$. 


I wouldn't be surprised if the outer automorphism groups of all finite groups of Lie type were known (they are not all known to me, but I'm no expert in these matters).


