Question

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 $Aut(G) \cong G$?

Answer 1

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 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).

Answer 2

The fact you are using above is really that all $S_n$, except for $n=2,6$, are complete groups. This means they are centerless, and all automorphisms are inner. Clearly, all complete groups $G$ are isomorphic to $Aut(G)$.

But the infinite dihedral group $D_\infty$ is centerless, yet not complete. We still have $D_\infty\cong Aut(D_\infty)$.

The dihedral group of order 8 $D_8$ is not centerless, and yet still $D_8\cong Aut(D_8)$.

I doubt very much there is a complete classification available.

Steve

Answer 3

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.