Given a prime $p$ and an integer $a$. If the order of $a$ modulo $p$ is 6, is it true that the order of $(1-a)$ modulo $p$ is also 6? If so prove it, if not give a counterexample.
It seems right. So I aimed to show that
$(1-a)^6 \equiv 1 \pmod p$, and also
$(1-a)^i \ne 1 \pmod p$ for all $i = 1, 2, 3, 4, 5$
Am I on the right track? Or is there any other way of thinking?