Let $\cal{U}$ be the unit group of group ring $\Bbb{Z}G$ then the Normalizer Problem (NP) states that $N_{\cal{U}}(G)=G\frak{z}$ where $\frak{z}=\cal{Z(U)}$.
Now why (NP) is equivalent to saying that $Aut_Z(G)=Inn(G)$ where $Aut_Z(G)$ denotes the automorphisms of $G$ induced by conjugation with units of $\Bbb{Z}G$.
If $u\in N_{\cal{U}}(G)$ then it certainly induces an automorphism of $G$ and if (NP) holds then conjugation by $u$ is certainly inner too. But I do not see how to deal with converse?
Thanks
Take an element $u \in N_{\mathcal{U}}(G)$, then we need to prove that $ u \in G \mathcal{Z}(\mathcal{U})$ (the other inequality is of course Always true). Now due to the definition of the normalizer we know that conjugation by $u$ on $G$ is an automorphism of $G$. Hence we find a $h \in G$ such that for any $g \in G$ we have: $$ u^{-1} g u = h^{-1}gh $$ We can rephrase this to find for any $g \in G$: $$ hu^{-1} g uh^{-1} = g $$ Thus we know that $uh^{-1} \in \mathcal{U}$ centralizes $G$, but then it also centralizes $\mathbb{Z}G$ (since $\mathbb{Z}$ is commutative). Hence we have that $uh{-1} \in \mathcal{Z}(\mathcal{U})$. Thus we find that $$ u = uh^{-1} h \in \mathcal{Z}(\mathcal{U}) G $$