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 $n$.

Yes, it is. Since $a \bar{a}=1$, it follows that for any positive integer $k$ we have $a^k (\bar{a})^k=1$. It follows that $a^k=1$ if and only if $(\bar{a})^k=1$. In particular, if $k$ is the smallest positive integer such that $a^k=1$, then $k$ is the smallest positive integer such that $(\bar{a})^k=1$. 
