I'm working in $ \mathbb{Z}/p \mathbb{Z} $ WLOG $m_p=0$ Assume that we cannot find such $k,l$ Then, $m_{\sigma(p)}=0$ and $m_{\sigma(1)}, m_{\sigma(2)}, ... , m_{\sigma(p-1)}$; $1m_{\sigma(1)}, 2m_{\sigma(2)}, ... , (p-1)m_{\sigma(p-1)}$ compile 2 complete residue systems. But then $-1=(p-1)!=1m_{\sigma(1)}*2m_{\sigma(2)}* ... *(p-1)m_{\sigma(p-1)}=(p-1)!m_{\sigma(1)}*m_{\sigma(2)}* ... * m_{\sigma(p-1)}=(p-1)!(p-1)!=1$. Absurd.

Very similar to: https://artofproblemsolving.com/community/c40244h1671238p10632173

This one's nice. First, WLOG $m_{i} = i$ for $i=1,...,p$. Now, assume for the sake of contradiction that this condition isn't satisfied for any pair $(i,j)$. This implies that $k\sigma{(k)}$ covers the entire residue class $\mod{p}$. Now, this implies that $\sigma{(p)}=p$. Now, we have $((p-1)!)^2=\prod_{n=1}^{p-1} n\sigma{(n)} \equiv (p-1)! \pmod{p}$, which is clearly a contradiction by Wilson's theorem. $\square$

The question is equivalent to " Let $a_1,a_2,..,a_p$ and $b_1,b_2,...,b_p$ be two reduced residue systems mod p where $p$ is prime. Let $c_i=a_i \cdot b_i$. Prove that $c_1,c_2,..,c_p$ is not a reduced residue system mod $p$ " Any product of elements of reduced residue system mod $p$ is equivalent to $-1$. But $$ \prod c_i= \prod a_i \cdot b_i= \prod a_i \cdot \prod b_i = -1 \cdot -1 =1.$$ Funny problem