Denote (still) $n\times n$ permutation matrices by $\mathfrak{S}_n$. The ordinary transpose preserves this group.

Given $P\in\mathfrak{S}_n$, construct the $n\times n$ matrix ${}^tP$ according to the rules:

(1) leave the 4 rims unchanged (1st row, 1st column, last row, last column);

(2) transpose its $(n-2)\times(n-2)$-submatrix found by removing the 4 rims.

Example. $$A=\begin{pmatrix} 1&0&0&0\\0&0&0&1\\0&1&0&0\\0&0&1&0 \end{pmatrix} \rightarrow {}^tA=\begin{pmatrix} 1&0&0&0\\0&0&1&1\\0&0&0&0\\0&0&1&0 \end{pmatrix}.$$ Example. $$B=\begin{pmatrix} 1&0&0&0\\0&0&0&1\\0&0&1&0\\0&1&0&0 \end{pmatrix} \rightarrow {}^tB=\begin{pmatrix} 1&0&0&0\\0&0&0&1\\0&0&1&0\\0&1&0&0 \end{pmatrix}.$$ Notice that ${}^tB$ is a permutation while ${}^tA$ is not!

Question. What is $a_n:=\#\{^{t}P\in\mathfrak{S}_n:\,\,P\in\mathfrak{S}_n\}$, for $\geq3$? Easy: $a_3=6$.

Remark. This is a modest case. We could easily generalize the question to transposing several submatrices.