Is there a non-trivial group G isomorphic to Aut(G)?

The title basically says it all.

Is there a group with more than one element that is isomorphic to the group of automorphisms of itself?

I'm mainly interested in the case for finite groups, although the answer for infinite groups would still be somewhat interesting.

• – Mariano Suárez-Álvarez Jul 2 '10 at 0:45
• Answer: yes. Smallest example: $S_3$, with order 6. Arguably most interesting example: monster simple group. – S. Carnahan Jul 2 '10 at 0:55
• From Mariano's link is the example of $D_8$, which is isomorphic to its automorphism group, but not by the natural map. – Ben Wieland Jul 2 '10 at 1:34

The automorphism group of the symmetric group $S_n$ is (isomorphic to) $S_n$ when $n$ is different from $2$ or $6$. In fact, if $G$ is a complete group you can ascertain that $G \simeq \mathrm{Aut}(G)$. The reverse implication needn't hold, though.
• As mentioned in the comments above, $D_8$ is a counterexample to the reverse implication, since it is not complete but is isomorphic to its automorphism group. – Joel David Hamkins Jul 2 '10 at 2:02
I am slightly surprised that Wielandt's automorphism tower theorem has not been mentioned: this asserts that given a finite group $G$ with trivial center, the sequence of groups defined by $G_{0} = G$ and $G_{n+1} = {\rm Aut}(G_{n})$ for $n \geq 0$ eventually stabilizes, that is at some stage ${\rm Aut}(G_{n}) = G_{n}$.
For the absolute Galois group of $\mathbb Q$, the map $g\mapsto (h \mapsto ghg^{-1})$ is an isomorphism between $G_\mathbb Q$ and its automorphism group. This is a corollary of the Neukirch-Uchida theorem.