Let $\{r_1, r_2, r_3, \ldots , r\phi(n)\}$ be a set of reduced residue system modulo $n$.
Show that $r_j^{−1} \equiv r_k \pmod n$ where $1 \le j, k \le \phi (n)$.
[The inverse of any residue in the reduced residue system is also in this system.]
By definition, $\gcd(r_j, n)=1$. This means we can use Euler's Theorem, which tells us
$$ r_j^{\phi(n)} \equiv 1 \pmod n $$
Factorising,
$$ r_j \times r_j^{\phi(n)-1} \equiv 1 \pmod n $$
And so the multiplicative inverse of $r_j$ is
$$r_j^{-1}=r_j^{\phi(n)-1}$$
Now, $\gcd(r_j,n)=1$ implies $\gcd(r_j^m,n)=1$ for any natural number $m$, including $m=\phi(n)=1$. That is
$$ \gcd(r_j^{\phi(n)-1}, n)=1 $$
This is membership criteria for being in the reduced residue system modulo $n$.
This means the inverse of any residue in the reduced residue system is also in the system.