Claim. Let $n$ be a natural number and $A=\{0,1,2,3,\cdots,n-1\}$ be a complete set of residues modulo $n$. Let $\sigma$ be a permutation of $A$. Show that the set $C=\{\sigma(i)i:i\in A\}$ is not a complete set of residues modulo $n$.
I can prove Claim for a prime number $n$. Indeed, suppose that $n>2$ and $C=\{\sigma(i)i:i\in A\}$ is a complete set of residues modulo $n$. Since $C$ has $n$ different elements, $\sigma(0)=0$. Then
$$\prod_{c\in C\setminus\{0\}}c =1\cdot 2\cdots (n-1)\equiv (n-1)! \equiv –1 \pmod n,$$
by Wilson's theorem. From the other side,
$$\prod_{c\in C\setminus\{0\}}c =\left(\prod_{a\in A\setminus\{0\}}a\right)^2\equiv (-1)^2 \equiv 1 \pmod n,$$
a contradiction!
But, unfortunately, I cannot prove the claim when $n$ is not a prime.