Let $A\leq \rm{Aut}(\mathbb{Z}/n\mathbb{Z})$. I am interested in the orbits of $\mathbb{Z}/n\mathbb{Z}$ under the action of $A$, i.e. the sets
$\{\sigma i : \sigma \in A\},$
where $i\in \mathbb{Z}/n\mathbb{Z}$. If $\sigma i = j$, then there exists a corresponding unit $u\in (\mathbb{Z}/n\mathbb{Z})^\times$ such that $ui\equiv j\bmod{n}$, whence $(i,n)=(j,n)$. Thus, to describe the behavior of the orbits of the group, it suffices to fix a divisor $d$ of $n$ and consider orbits of the set
$R_d:=\{i\in\mathbb{Z}/n\mathbb{Z} : (i,n)=\tfrac{n}{d}\},$
which has $\varphi(d)$ elements. Is anything known about how an arbitrary automorphism subgroup of $\mathbb{Z}/n\mathbb{Z}$ partitions $R_d$ into orbits? In particular, can anything be said about their size and number?
Edit:
Perhaps I should clarify. The fact that $\rm{Aut}(\mathbb{Z}/n\mathbb{Z})\cong (\mathbb{Z}/n\mathbb{Z})^\times$ was used above to deduce that each orbit of $\mathbb{Z}/n\mathbb{Z}$ is contained in some $R_d$. I am interested in the explicit nature of the orbits. For example, if $A$ is cyclic, then I have shown that $A$ partitions $R_n$ into $|A|$ orbits of size $\varphi(n)/|A|$. It would be nice to generalize this sort of result to include cases where $d< n$ and $A$ is not cyclic.