Show that if $a^{−1} ≡ b \pmod n$ then $b^{−1} ≡ a \pmod n$.
We're given $a^{−1} ≡ b \pmod n$ which means
$$ab \equiv 1 \pmod n$$
Since $ab = ba$, we have
$$ba \equiv 1 \pmod n$$
Which gives us that $a \pmod n$ is the multiplicative inverse of $b$ modulo $n$. That is, $b^{-1} \equiv a \pmod n$.