Let $A$ be an $n\times n$ matrix over a field, and consider the matrix $B$ obtained by swapping two diagonals of $A$, where the diagonals "wrap around" like in the calculation of the determinant of a $3\times 3$. For example, for $n = 3$, one valid swapping would be:
$$B = \begin{pmatrix} \color{blue}c & b & \color{red}a\\ \color{red}e & \color{blue}d & f\\ g & \color{red}i & \color{blue}h\end{pmatrix}\text{ where }A = \begin{pmatrix} \color{red}a & b & \color{blue}c\\ \color{blue}d & \color{red}e & f \\ g &\color{blue} h & \color{red}i\end{pmatrix}$$
Can this operation change the rank of a matrix?
It's easy to see that for $n = 2$ it doesn't, as it simply interchanges and transposes the set of rows and comluns of $A$. It seems that this is also the case for general $n \times n$, but I still coundn't put my finger down on a proof.
EDIT: One can even make the swapping cannonical if we assume it also preserves the antidiagonals. That is, given two diagonals $D$ and $D'$ inside $A$, we swap an element $a$ in $D$ with the unique element $b$ in $D'$ that lies in the same antidiagonal as $a$. The $n = 3$ example then becomes:
$$B = \begin{pmatrix} \color{blue}h & b & \color{red}e\\ \color{red}i & \color{blue}c & f \\ g & \color{red}a & \color{blue}d\end{pmatrix}\text{ where }A = \begin{pmatrix} \color{red}a & b & \color{blue}c\\ \color{blue}d & \color{red}e & f \\ g &\color{blue} h & \color{red}i\end{pmatrix}$$
I'm particularly interested in just the existence of a rank preserving swap, so this canonical one has more chance of working (eg. in the $3\times 3$ case it preserves invertibleness).