Switching rows and switching columns in an $n\times n$ chessboard leaves the board unchanged if the first row and first column are unchanged Consider an $n\times n$ matrix, over $\mathbb{F}_2$, arranged in a chessboard like manner: the neighbors of a $1$ are all $0$'s, and vice versa. Here's an example with $n=3$
$$
\begin{bmatrix}
1 & 0 & 1\\
0 & 1 & 0\\
1 & 0 & 1
\end{bmatrix}
$$
We'll say such a matrix is ordered.
Define a shuffle of such a matrix to be the resulting matrix after a finite number of row switching $R_i \leftrightarrow R_j$, and column switching $C_i \leftrightarrow C_j$ have been applied to an ordered matrix.
It seems reasonable that if a shuffle has both its first row and its first column in order, then the matrix is ordered (likewise for any row and column). I need help in proving it, or finding a counter-example.
 A: If the same permutation is applied to identical sequences, they remain identical; if the same permutation is applied to complementary sequences, they remain complementary. So permuting columns doesn't change which rows are identical and which are complementary; and permuting rows doesn't change which columns are identical and which are complementary.
Furthermore, two rows are identical precisely when their first-column elements are the same, and complementary precisely when their first-column elements differ. The analogous statements hold for columns.
So if, after permutation of rows and columns, row $1$ looks like
$$
\begin{bmatrix} 1 & 0 & 1 & 0 & 1 & \ldots\end{bmatrix}
$$
then columns $3,\ 5,\ 7, \ldots$ are identical to column $1$ and columns $2,\ 4,\ 6,\ \ldots$ are complementary to it. And if column $1$ looks like
$$
\begin{bmatrix} 1 & 0 & 1 & 0 & 1 & \ldots\end{bmatrix}^T
$$
then rows $3,\ 5,\ 7, \ldots$ are identical to row $1$ and rows $2,\ 4,\ 6,\ \ldots$ are complementary to it.
