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 $Aut(G) \cong G$?
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Sign up to join this communityIn 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 $Aut(G) \cong G$?
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).
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
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.