Problem: Given a cyclic group $ C=\langle c\rangle$ of order $ n $ acting on a finite set $ X $, let $$ B\left(d\right)=\#\{x \in X:x \text{ is fixed by at least the subgroup } C_d=\langle c^{n/d}\rangle \text{ of } C\} $$ Then the number of $ C $-orbits is $$ \frac 1 n \sum_{d \mid n}{\varphi(d)B(d)} $$
My approach: The number of orbits is given by the following formula: $$ \frac 1 n \sum_{g \in G}{\# X^g} $$ Since $ \varphi(d) $ calculates the number of elements of order $ d $, the conclusion follows.
Question: It seems that we can also get the answer by using Mobius inversion. But I can't come up with any solution. Need help QQ