From: "Dr. Robert Baddeley" 
Date:          Mon, 17 Mar 1997 10:53:52 GMT0BST
Subject:       Re: Automorphisms of a wreath product
 
Dear Jaoa, 
 
Let $G$ be a group, and let $P$ be a transitive
permutation group on $\Omega$  (possibly 
infinite). Let $W$ be the restricted wreath product $G\wr P$. 
Denote the base group by $B$ (which is a direct product of copies of 
$G$).
 
If $B$ is a characteristic subgroup of $W$, then we can view $Aut W$
as the normalizer in $Aut B$ of $W$. So we are interested in when $B$ 
is not characteristic. Fletcher Gross, in a paper `On the uniqueness 
of wreath products' J. Algebra (which has appeared but I'm not sure 
when), proved that if one of the following holds:
 
(i) $P$ is not isomorphic to any proper factor group of itself;
(ii) for some positive integer $n$ the $n^{th}$ term $G^{(n)}$
of the derived series of $G$ is not isomorphic to a proper subnormal 
subgroup of itself;
(iii) the stabilizer in $P$ of an element of $\Omega$ has finite 
order;
then
$B$ is not characteristic subgroup only if $P$ has a non-identity
normal elementary abelian 2-subgroup and $G=\langle u\rangle H$
where $H$ is abelian, $u^2=1$, and $u^{-1} h u = h^{-1}$ for all 
$h\in H$.                      
 
Gross says more in the finite case. His results have been extended a 
little by Laci Kov\'acs and myself (as yet unpublished). 
 
If $W$ is finite wtih $P\le S_n$  and $B$ is not characteristic, then $G=
\langle u\rangle H$ with $H$ abelian of odd order, $u^2=1$,
$h^u=h^{-1}$ for all $h\in H$, and $P$ contains a normal
elementary abelian 2-group of size $2^{n/2}$; 
moreover the $Aut W$ orbit of $B$
has length 2 precisely. 
 
 
In particular, if $P$ is the full symmetric group then $B$
is characteristic!!
 
Cheers, 
Robert.