Consider the group of $mn\times mn$ permutation matrices $\mathfrak{S}_{mn}$ and partition each such matrix $P$ into $n^2$ blocks of $m\times m$ matrices $Q_{i,j}$. Now, transpose each $Q_{i,j}$ (independently) to form a new $mn\times mn$ matrix denoted $P^t$ (with an abuse of notation). Let's construct the set $U_{mn}:=\{P^t\in\mathfrak{S}_{mn}:\, P\in\mathfrak{S}_{m,n}\}$.

It's clear that if $m=1$ then $U_{mn}=\mathfrak{S}_{mn}$; the same if $n=1$.

Question 1.For which $m$ and $n$, does $U_{mn}$ form a group?UPDATE. Negative answer shown below.

A cute note: $\frac{\#\mathfrak{S}_{2n}}{\# U_{2n}}=C_n$ is the Catalan number. Is $\# U_{2n}$ a subgroup?
**Answer.** No. See below. Another counterexample to the converse of Lagrange's theorem: $\# U_{2n}$ divides $\#\mathfrak{S}_{2n}$ but $U_{2n}$ is not a subgroup of $\mathfrak{S}_{2n}$.

Question 2.View $P\in U_{2n}$ as $P\leftrightarrow\sigma$ as a $1$-line permutation $\sigma=(\sigma_1,\dots,\sigma_{2n})$. In this way, what is an equivalent (to the above "transpose") characterization of $P$ in the language of $\sigma$?UPDTE. I am Still waiting for an answer.

If you're interested in enumeration then visit here.