Can transposition $GL_n(k)\to GL_n(k)$ ever be realized as conjugation by some element in $S_n$?

Question

Can the transpose operation on $GL_n(k)$ ever be expressed as conjugation by some element of $S_n$, identified as the group of permutation matrices? That is, does there exist $\sigma\in S_n$ such that $A^T=\sigma A\sigma^{-1}$ for all $A$?

I'm doubtful, because the tranpositions in $S_n$ act on the right by swapping columns, and act on the left by swapping rows, hence any permutation matrix does some sequence of such swaps. So I don't see a way to flip a column across the main diagonal using just these types of moves, since all the entries in a given column will stay together (as possibly a different column, and in a different "vertical" order after swapping rows around.)

Answer 1

To see why this is impossible, consider a matrix all of whose entries are distinct, for example $A=(2^i(2j+1))_{i,j}$.

If $B=\sigma A\sigma^{-1}$ leaves $a_{ii}$ unchanged (i.e. $b_{ii}=a_{ii}$), this means that $\sigma$ fixes $i$.

So if $\sigma A\sigma^{-1}$ leaves the diagonal unchanged, then $\sigma$ must be the identity.

Answer 2

Since $A^t=A$ whenever $A$ is diagonal, such $\sigma$ would commute with all diagonal matrices in $GL_n(k)$. When $|k|>2$, you can easily see that $\sigma$ needs to be the identity, proof :

Let $\sigma\in\mathfrak{S}_n$ be different from identity. Let $j>i$ such that $\sigma(i)=j$. Define the matrix $A\in GL_n(k)$ which is diagonal with all diagonal elements being equal to one except the $i$-th (this where we need two invertible elements in $k$). Then $\sigma\cdot A\neq A$.

That leaves you with the case $k=\mathbb{F}_2$. I think you can compute counter-example for this field for any $n>1$.