Claim: if $a_1,a_2,\dots,a_k$ is a reduced residue system modulo $n$ then $aa_1,aa_2,\dots,aa_k$ is also a reduced residue system modulo $n$.
proof:
Given $\gcd(a,n)=1$. We can find an $x$ such that $ax\equiv 1\pmod{n}$. Also for $a_i$ in the range $1\leq i\leq k$ we have $\gcd(a_i,n)=1$ by definition. So we can find a $y_i$ such that $a_iy_i\equiv 1\pmod{n}$. Then $(xy_i)(aa_i)\equiv(ax)(y_ia_i)\equiv 1\pmod{n}$. This proves that $aa_i$ is invertible modulo $n$. Hence $\gcd(aa_i,n)=1$, that is the $aa_i$'s are coprime to $n$. If $aa_i\equiv aa_j\pmod{n}$, then $xaa_i\equiv xaa_j\pmod{n}$, or $a_i\equiv a_j\pmod{n}$. Since the $a_i$'s are distinct modulo $n$, this is only true if $i=j$. Hence the $aa_i$'s are also distinct modulo $n$. Therefore, $aa_1,aa_2,\dots,aa_k$ is also a reduced residue system modulo $n$.
Is my proof correct? I am unsure about the step where I regroup the product, that is if that actually shows $aa_i$ is invertible or not.
You just need to show $|\{aa_1,aa_2,\dots,aa_k\}|=k$. Suppose it is less than $k$. Then exist $a_i,a_j$, $i\ne j$, such that $$aa_i\equiv aa_j\pmod n$$
But then $n\mid a(a_i-a_j)$ so, by Gauss lemma ($n$ and $a$ are relatively prime) we have $n\mid a_i-a_j$, but this means $a_i\equiv a_j\pmod n$. A contradiction.