Consider $n$ permutations $(\pi_i; 1 \le i \le n)$ of $[k]$. Suppose that for any transposition $\sigma = (j, l)$, $1 \le j, l \le k$, \begin{equation} \label{1} (*) \quad \sum_{i = 1}^n inv(\pi_i \circ \sigma) > \sum_{i = 1}^n inv(\pi_i), \end{equation} where $inv(\pi_i)$ is the number of inversions of $\pi_i$. Is it possible that \begin{equation} (**) \quad \sum_{i = 1}^n inv(\pi_i \circ \pi) \le \sum_{i = 1}^n inv(\pi_i) \quad \mbox{for some permutation } \pi \neq id. \end{equation} The answer is no for n = 1 and 2, since $inv(\pi_i \circ \sigma) = inv(\pi) \pm 1$ for any transposition $\sigma$. In these cases, $(*)$ implies that $\pi_i = id$ and $(**)$ is automatically true. The answer also seems to be negative for $\pi$ a permutation of adjacent three elements, e.g. $(2,3,1)$. I want to know if any positive or negative result holds in general.