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?


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.


Since $\Bbb Z/n\Bbb Z$ is cyclic, any automorphism of $\Bbb Z/n\Bbb Z$ is completely determined by its effect on a fixed generator (e.g. on $\bar1$). On the other hand, it is clear that for any $\bar k\in(\Bbb Z/n\Bbb Z)^\times$, i.e. any class of additive period $n$, the assignment $\bar1\mapsto\bar k$ defines an automorphism $\phi_{\bar k}$.

It is now completely straightforward to check that the map $$ (\Bbb Z/n\Bbb Z)^\times\longrightarrow\text{Aut}(\Bbb Z/n\Bbb Z),\qquad \bar k\mapsto\phi_{\bar k} $$ is an isomorphism, from which you can easily get the answers you seek.

  • $\begingroup$ Perhaps it would also be interesting to note that under the Chinese Remainder theorem, $(\mathbb Z / n \mathbb Z)^{\times} \cong \prod_{p \, | \, n} (\mathbb Z / p^{\alpha_p} \mathbb Z)^{\times}$, which gives a more explicit expression for $\mathrm{Aut}(\mathbb Z / n\mathbb Z)$. Good answer though, +1 $\endgroup$ – Patrick Da Silva Aug 12 '12 at 7:45

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