TITLE: $16$ rooks on a chess board.
QUESTION [6 upvotes]: Place $8$ pairwise non-attacking white rooks and black rooks on a $8\times8$ chess board. If one can swap rows and columns, is it possible for the black rooks to take the initial position of the white rooks and vice versa?
My attempt:
Let the rows be $a_1,a_2,...,a_8$ and columns $b_1,b_2,...,b_8$. If there are white rooks at ($a_i$,$b_j$) and ($a_j$,$b_a$) as well as black rooks at ($a_i$,$b_i$) and ($a_j$,$b_j$), then we can reduce it to a $2\times2$ and a $6\times6$. 
I don't know what to do next. Any help appreciated.

REPLY [5 votes]: The answer is yes, it is possible. It is in fact true on any $n\times n$ chessboard. I have a solution which is easy enough to understand provided you know a little group theory. In particular, you need to be comfortable with the symmetric group. 
Number the rows and columns of the board $1,2,\dots,n$. For every $i$, there is a single white rook in row $i$; let $\pi_i$ be the column of that rook. Similarly, let $\rho_i$ be the column of the black rook in row $i$. Then $\pi$ and $\rho$ are permutations. 
We are allowed to permute the rows and columns of the board. Let $\sigma$ and $\tau$ represent the permutations of the rows and columns. These are valid solutions if and only if
$$
\sigma\pi\tau=\rho\qquad \text{and}\qquad \sigma \rho \tau=\pi
$$
Solving the first equation for $\tau$, and plugging that expression into the second equation, we get
$$
\sigma\rho\pi^{-1}\sigma^{-1}\rho=\pi
$$
or
$$
\sigma(\pi\rho^{-1})^{-1}\sigma^{-1}=\pi\rho^{-1}\tag1
$$
Equation $(1)$ is satisfiable if and only if the permutation $\pi\rho^{-1}$ is conjugate to its inverse. But every permutation is conjugate to its inverse, as any permutation $\alpha$ and its inverse $\alpha^{-1}$ have the same cycle structure. The inverse of a permutation is found by reversing the elements of each cycle, which does not affect the structure.