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?



Is ${\rm S}_6$ the automorphism group of a group?
${\rm S}\_6$ 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](http://dx.doi.org/10.1216/RMJ-1979-9-4-653).


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 $X$.


