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 beganthe study of $\rm{Aut}_{sn}(G)$. Other authors have also considered thestructure of $\rm{Aut}_{sn}(G)$ under various restrictions on thestructure of $G$ (Robinson [11], Cossey [2], DalleMolle [4]). The inner automorphisms in $\rm{Aut}_{sn}(G)$ areprecisely the inner automorphisms induced by elements of $\omega(G)$,the Wielandt subgroup of $G$. Recall that the Wielandt subgroup of agroup $G$ is the set of all elements of $G$ that normalise eachsubnormal subgroup of $G$ and that $\zeta(G)$, the centre of $G$, iscontained in $\omega;(G)$. Thus $\rm{Aut}_{sn}(G)\cap;\rm{Inn}(G)$ isisomorphic to $\omega(G)/\zeta(G)$ and some of the results obtainedindicate that the structure of $\rm{Aut}_{sn}(G)$ is controlled by thestructure of $\omega(G)/\zeta(G)$; for example, Robinson [11,Corollary 3] shows that, for a finite group $G,\rm{Aut}_{sn}(G)$ isinsoluble if and only if $\omega(G)$ is insoluble. We shall prove aresult of a similar nature here. One of the main results (Theorem B) ofFranciosi 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.