Let $k$ be the order of $a(\mathrm{mod}n)$, then $a^b\equiv 1(\mathrm{mod}n)$ if and only if:

By definition, the order $k$ of an element $a(\mathrm{mod}n)$ is the smallest positive integer such that $a^k\equiv 1(\mathrm{mod}n)$.A fundamental property of the order is that for any integer $b$, $a^b\equiv 1(\mathrm{mod}n)$ if and only if the order $k$ divides the exponent $b$.If $b=kq+r$ (where $0\le r<k$), then:
$$a^b=a^{kq+r}=(a^k{)}^q\cdot a^r\equiv 1^q\cdot a^r\equiv a^r(\mathrm{mod}n)$$
For $a^b\equiv 1(\mathrm{mod}n)$ to hold, we must have $a^r\equiv 1(\mathrm{mod}n)$. Since $k$ is the smallest such positive integer and $r<k$, $r$ must be $0$. Thus $k$ divides $b$.