Inversion inequalities

Question

Do there exist $n$ permutations $(\pi_i; 1 \le i \le n)$ of $[k]$ such 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 $\pi_i$, but \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 adjacent 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.

Answer 1

The smallest $k$ when such permutations exist is $k=4$. Namely, the set of $\pi_i$ given by $$\big\{\ [1, 2, 3, 4], [1, 3, 4, 2], [1, 4, 2, 3], [2, 1, 3, 4], [2, 3, 4, 1], [2, 4, 1, 3],$$ $$ [3, 4, 1, 2], [3, 4, 2, 1], [4, 1, 2, 3], [4, 1, 3, 2], [4, 2, 1, 3]\ \big\}$$ and $\pi = [2, 3, 4, 1]$ do the job. In fact, the total number of inversions in this set is decreased by 5 when the elements are multiplied by $\pi$, while multiplying them by any transposition increases the total number of inversions by at least 1.

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ADDED. A smaller set example for the same $\pi = [2, 3, 4, 1]$: $$\{ [1, 2, 3, 4], [3, 4, 1, 2], [4, 1, 2, 3] \}.$$