Is the multiplicative order of a number always equal to the order of its multiplicative inverse? Is it true that $ord_{n}(a)=ord_{n}(\bar{a})$ $\forall n$?
Here, $\bar{a}$ refers to the multiplicative inverse of $a$ modulo $n$ and $ord_{n}(a)$ refers to the multiplicative order of $a$ modulo $n$.

The answer is yes.
Hint: note that $a^m (\bar a)^n = a^{m-n}$.
