Is ${\rm S}_6$ the automorphism group of a group? The automorphism group of the symmetric group $S_n$ is $S_n$ when $n$ is not $2$ or $6$, in which cases it is respectively $1$ and the semidirect product of $S_6$ with the (cyclic) group of order $2$. (For this famous outer automorphism, see for instance wikipedia or Baez's thoughts on the number $6$.)
On the other hand, $S_2$ is the automorphism group of $Z_3$, $Z_4$ and $Z_6$ (and only those groups among finite groups). Hence my question: is $S_6$ the automorphism group of a group? of a finite group?
 A: ${\rm S}_6$ is not the automorphism group of a finite group.
See H.K. Iyer, On solving the equation Aut(X) = G, Rocky Mountain J. Math. 9 (1979), no. 4, 653--670, available online
here.
This paper proves that for any finite group $G$, there are finitely many 
finite groups $X$ with ${\rm Aut}(X) = G$, and it explicitly solves the
equation for some specific values of $G$.
In particular, Theorem 4.4 gives the complete solution for $G$
a symmetric group, and when $n = 6$ there are no such $X$.
A: There is a whole array of results, going back to G. Birkhoff at 1930s saying that every group is an automorphism group of some universal algebra (or some universal algebra inside some class).

(This really should be merely a comment to the previous answer, but I am still not reputable enough to leave comments).
A: It's probably worth pointing to

Belolipetsky, Mikhail; Lubotzky,
  Alexander.  Finite groups and
  hyperbolic manifolds. Invent. Math.
  162 (2005), no. 3, 459-472.   MR2198218.

where it is shown that for every finite group G, there's an infinite group Gamma with Out(Gamma) = G.
A: On the other hand, ${\rm S}_6$ is isomorphic to ${\rm Sp}_4(\mathbb{F}_2)$,
so that it is an automorphism group in another category (other than groups
or sets). This automorphism is exhibited by looking at the 2-torsion of the 
Jacobian of a hyperelliptic curve H of genus 2 (if $H$ is given by
$y^2 = f(x)$, with $f$ of degree 6, then 15 non-trivial two torsion points are 
given [as a Galois module] by differences of roots of $f$; see the wiki page 
for Kummer surface).
