An $n\times n$ matrix that has exactly one $1$ and one $-1$ in each row and column and others are $0$ I came across the following Question

Assume an $n\times n$ matrix that has exactly one $1$ and one $-1$ in
each row and column and others are $0$. Prove that there is a way that
we can change the places of rows and columns in which it gives the
negative of the matrix.

MY TRY-
Call such matrix A.
All we need to do is to find some permutation matrices $P_{1}$ and $P_{2}$ such that $$P_{1}AP_{2} = -A$$
$A$ can be written as a difference of two permutation matrices i.e. $$A = P-Q$$
where P and Q are some permutation matrices
Example of one such matrix of order $3\times3$
$$ \begin{pmatrix}
1 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 1
\end{pmatrix} = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
 0 & 0 & 1
\end{pmatrix}-\begin{pmatrix}
0 & 0 & 1 \\
1 & 0 & 0 \\
 0 & 1 & 0
\end{pmatrix}.$$
We could first turn every such matrix $A$ by multiplying
by appropriate permutation matrices to the form $I-R$ :-$$P^{T}A = P^{T}(P-Q) = I-R$$ Clearly the permutation matrix R shouldn't have $1$ at the same position as in $I$. R lies in the class of traceless permutation matrices. Now If we are able to find matrices permutation $P_{1}$ and $P_{2}$ such that $$P_{1}(I-R)P_{2} = (R-I) = -(I-R)$$ we'll have
$$P_{1}P^{T}AP_{2} = -P^{T}A \implies PP_{1}P^{T}AP_{2} = -A $$
and we would be done.
But how could I proceed now to find $P_{1}$ and $P_{2}$?
Would we need some extra equation from the fact that $R$ is a traceless permutation matrix?
It was great to see other approaches to solve the problem by Michael Hoppe and user1551.
But I am curious to see how would it be if we go this way?
 A: Not a solution, but a direction in which to go
Your idea of "difference of permutations" is a nice one for describing these "good" matrices, but as you observe, it doesn't, in its current form, seem to be leading you anywhere.
You've said that not every difference of permutations is "good", and that's true. And you want to find a property that's characterizes the ones that are good. And you've actually identified the property: they never have a "1" in the same position.
Now if you have a difference of permutations that's "good", and you left-multiply by a permutation, you STILL have a difference of permutations, i.e., $P_1(P-Q) = (P_1P) - (P_1 Q)$. The only question is "do the matrices $P_1P$ and $P_1Q$ still have the "no 1s in the same position" property.
(You then have to do the same thing for right-multiplying, but that'll be easy of the left-multiply thing works out).
So here's a lemma to prove:
If $A, B, P$ are permutations, and $A$ and $B$ have no $1$s in the corresponding positions, then $PA$ and $PB$ have no $1$s in corresponding positions either.
That should get you going.
A: As said in the question, it is enough to work with $A=I-R$, where $R$ is a permutation matrix without 1 on the diagonal. Suppose that $R$ is the matrix of the permutation $p$. We will show later that every permutation is a product of two involutions, that is, we can write $p=fg$ where $f^2=g^2=id$. (Here a product $fg$ maps $i$ to $f(g(i))$ for all $i$). If $F,G$ are the matrices corresponding to $f,g$, then we have
$R=FG$ and $F^2=G^2=I$. Then the statement follows from the fact that
$$F(I-R)G=FG-F^2G^2=R-I=-(I-R).$$
It remains to show that every permutation $p$ is the product of two involutions. We can write $p=c_1c_2...c_k$ as a product of disjoint cycles $c_j$ (see here). Therefore it is sufficient to write cycles as a product of two involutions. Furthermore it is sufficient to do this for the cycle corresponding to the mapping $c:i\mapsto i+1 \mod m$. Here we can write it as a product $c=fg$ where $f:i\mapsto m+1-i\mod m$ and $g:i\to m-i \mod m$.
More explicitly, a cycle $c=(a_1\,a_2\,\dots\,a_m)$ is the product $c=fg$ of the involutions
$$f=\begin{pmatrix}a_1&a_2&\dots &a_m\\a_m&a_{m-1}&\dots &a_1\end{pmatrix}
\mbox{ and }g=\begin{pmatrix}a_1&a_2&\dots&a_{m-1} &a_m\\a_{m-1}&a_{m-2}&\dots &a_1&a_m\end{pmatrix}.$$
The factorisations of different cycles in the product $f=c_1c_2\dots c_k$ do not interfere with each other as they concern different sets.
This completes the proof.
A: Here's an algorithm to transform the matrices.  I'll explain by an example.
We want to transform
$$
\begin{pmatrix}
-1 & 0 & 1 & 0\\
0 & -1 & 0 & 1\\
0 & 1 & -1 & 0\\
1 & 0 & 0 & -1
\end{pmatrix}\quad\text{to}\quad
\begin{pmatrix}
1 & 0 & -1 & 0\\
0 & 1 & 0 & -1\\
0 & -1 & 1 & 0\\
-1 & 0 & 0 & 1
\end{pmatrix}.
$$
We may define the companion of the first matrix as
$$\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 1 & 2
\end{pmatrix}$$
where the companion's first column gives the position of $-1$ and $1$ in the first column of the matrix resp., that is $\left(\begin{smallmatrix}1\\4\end{smallmatrix}\right)$ and so on.
Now changing two columns in the matrix changes the correspondent columns in the companion; exchanging two rows $j$ and $k$ in the matrix exchanges all values of $j$ and $k$ in the companion.
We want to go from
$$\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 1 & 2
\end{pmatrix}\quad\text{to}\quad\begin{pmatrix}
4 & 3 & 1 & 2\\
1 & 2 & 3 & 4
\end{pmatrix}.$$
Start with exchanging the first and last column of the companion:
$$\begin{pmatrix}
4 & 2 & 3 & 1\\
2 & 3 & 1 & 4
\end{pmatrix}.$$
Now the first column should be $\left(\begin{smallmatrix}4\\1\end{smallmatrix}\right)$, hence we swap row $1$ and $2$ to obtain
$$\begin{pmatrix}
4 & 1 & 3 & 2\\
1 & 3 & 2 & 4
\end{pmatrix}.$$
Repeat the process with the companion's second column, that is switch columns $2$ and $3$:
$$\begin{pmatrix}
4 & 3 & 1 & 2\\
1 & 2 & 3 & 4
\end{pmatrix}$$
and we're done already. That was easily done, now another example for the systematic way
Take the companion
$$\begin{pmatrix}
2 & 3 & 1 & 5 & 4\\
4 & 5 & 3 & 1 & 2
\end{pmatrix}.
$$
Look for cycles in the permutation, there are two, namely $(3,5,1)$ and $(2,4)$.  Now first change $3$ to $5$, that is, exchange row $3$ with row $5$ in the corresponding matrix to get
$$\begin{pmatrix}
2 & 5 & 1 & 3 & 4\\
4 & 3 & 5 & 1 & 2
\end{pmatrix},
$$
then exchange $5$ and $1$:
$$\begin{pmatrix}
2 & 1 & 5 & 3 & 4\\
4 & 3 & 1 & 5 & 2
\end{pmatrix}
$$
and for the first cycle finally $1$ and $3$:
$$\begin{pmatrix}
2 & 3 & 5 & 1 & 4\\
4 & 1 & 3 & 5 & 2
\end{pmatrix}.
$$
For the second cycle exchange $2$ and $4$
$$\begin{pmatrix}
4 & 3 & 5 & 1 & 2\\
2 & 1 & 3 & 5 & 4
\end{pmatrix}.$$
Now change the columns for the correct order:
$$\begin{pmatrix}
4 & 5 & 3 & 1 & 2\\
2 & 3 & 1 & 5 & 4
\end{pmatrix}.$$
Done!
For the first example we could have perform the changes $1\leftrightarrow4$, $4\leftrightarrow2$, $2\leftrightarrow3$, and $3\leftrightarrow1$ and then switch the columns accordingly, but there obviously was an easier way.
