We denote by $\rm{Aut}_{sn}(G)$
the set of all automorphisms that fix every subnormal subgroup of $G$
setwise. In their paper [5], Franciosi and de Giovanni began
the study of $\rm{Aut}_{sn}(G)$. Other authors have also considered the
structure of $\rm{Aut}_{sn}(G)$ under various restrictions on the
structure of $G$ (Robinson [11], Cossey [2], Dalle
Molle [4]). The inner automorphisms in $\rm{Aut}_{sn}(G)$ are
precisely the inner automorphisms induced by elements of $\omega(G)$,
the Wielandt subgroup of $G$. Recall that the Wielandt subgroup of a
group $G$ is the set of all elements of $G$ that normalise each
subnormal subgroup of $G$ and that $\zeta(G)$, the centre of $G$, is
contained in $\omega;(G)$. Thus $\rm{Aut}_{sn}(G)\cap;\rm{Inn}(G)$ is
isomorphic to $\omega(G)/\zeta(G)$ and some of the results obtained
indicate that the structure of $\rm{Aut}_{sn}(G)$ is controlled by the
structure of $\omega(G)/\zeta(G)$; for example, Robinson [11,
Corollary 3] shows that, for a finite group $G,\rm{Aut}_{sn}(G)$ is
insoluble if and only if $\omega(G)$ is insoluble. We shall prove a
result of a similar nature here. One of the main results (Theorem B) of
Franciosi and de Giovanni [5] is that, for a polycyclic group
$G$, $\rm{Aut}_{sn}(G)$ is either finite or abelian. We shall show that
$\rm{Aut}_{sn}(G)$ can indeed be infinite, but only if
$\omega(G)/\zeta(G)$ is infinite.