Action on $\mathbb{Z}_n$

Question

Let $2^n - 1 = rt$. Сonsider map $\phi : \mathbb{Z}_r \rightarrow \mathbb{Z}_r$ such that $\phi(a) = 2a$. The question is how many orbits this action has? Clearly we can see that if $a \equiv a2^m \text{ mod } r$ so orbit's order is $ m-1 = |\mathrm{Orb}(a)| \leq n $ because $r \mid 2^n - 1$. And there is only one orbit that has order 1, the others has order $2^k$. Nothing more I can say.