All permutation matrices that convert one Hadamard matrix into another Hadamard matrix. Given a Hadamard matrix $H$, I know that applying row and column permutations, along with multiplying a row or a column with a -1 results in another Hadamard matrix $H^{'}$ equivalent to the first. 
Given $H$ and $H^{'}$, ( assume we know that they're equivalent) , is it possible to find all pairs of permutation matrices $(P,Q)$ that preserve Hadamard equivalence, such that,
\begin{equation}
H^{'}=PHQ
\end{equation}
?
For example, take 2 Hadamard matrices of order 4, $H,H^{'}$. We know that there is only one equivalence class of Hadamard matrices of order 4, so the 2 matrices are equivalent. 
$H =
\left(
\begin{matrix}
1 & 1 & - & 1 \\
- & 1 & - & -\\
1 & 1 & 1 & -\\
- & 1 & 1 & 1 \\
\end{matrix}
\right)
$, 
$H^{'} =
\left(
\begin{matrix}
1 & 1 & 1 & - \\
- & 1 & 1 & 1\\
- & - & 1 & -\\
1 & - & 1 & 1 \\
\end{matrix}
\right)
$
Now, 
$
\left(
\begin{matrix}
1 & 1 & 1 & - \\
- & 1 & 1 & 1\\
- & - & 1 & -\\
1 & - & 1 & 1 \\
\end{matrix}
\right)
=$
$
\left(
\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1 \\
\end{matrix}
\right)
*$
$
\left(
\begin{matrix}
1 & 1 & - & 1 \\
- & 1 & - & -\\
1 & 1 & 1 & -\\
- & 1 & 1 & 1 \\
\end{matrix}
\right)
*$
$
\left(
\begin{matrix}
0 & 0 & 0 & - \\
0 & 0 & 1 & 0\\
0 & - & 0 & 0\\
1 & 0 & 0 & 0 \\
\end{matrix}
\right)
$
So here,
$P =
\left(
\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1 \\
\end{matrix}
\right)
$
,
$Q =
\left(
\begin{matrix}
0 & 0 & 0 & - \\
0 & 0 & 1 & 0\\
0 & - & 0 & 0\\
1 & 0 & 0 & 0 \\
\end{matrix}
\right)
$
This is one example of a pair $(P,Q)$. How do you list all such pairs?
(I'll make the question more specific and add that I'm looking for permutation matrices for 4x4 Hadamard matrices only)
 A: Let $P$ and $Q$ be monomial matrices such that $H'=PHQ,$ and let $P_1$ and $Q_1$ be some other set of monomial matrices such that $H'=P_1HQ_1.$  Then, since monomial matrices form a group, there are monomial matrices $R$ and $S$ such that $P_1=RP$ and $Q_1=QS.$  So $H'=PHQ=RPHQS=RH'S,$ and therefore $(R,S)$ is an automorphism of $H'.$
Conversely, let $(R,S)$ be an automorphism of $H'.$  Then if $H'=PHQ$ for some pair of monomial matrices $(P,Q),$ we also have $H'=(RP)H(QS).$
As a consequence, we can answer your question if we can list all of the automorphisms of $H'.$
If $(R,S)$ is an automorphism of $H',$ then $S$ is uniquely determined by $R$ since $H'$ is nonsingular.  The automorphism group of any $\pm1$ matrix always contains two trivial automorphisms: $(R,S)=(I,I)$ and $(R,S)=(-I,-I).$  The automorphism group of a $4\times4$ Hadamard matrix, however, is nontrivial.
Since all $4\times4$ Hadamard matrices are equivalent, we only need to compute the automorphism group for one particular matrix.  A nice choice is
$$
H'=\begin{bmatrix}
-&1&1&1\\
1&-&1&1\\
1&1&-&1\\
1&1&1&-
\end{bmatrix}.
$$
This matrix has one element $-1$ in each row and column.  Observe that negating a row of $H'$ changes the parity of the number of elements $-1$ in all of the columns of $H',$ but does not change the parity of the the number of elements $-1$ in any row of $H'.$  Likewise, negating a column does not change the parity of the number of elements $-1$ in any column.  Therefore, in order that $(R,S)$ be an automorphism of $H',$ that is, in other that $RH'S=H',$ it is necessary that the number of elements $-1$ in $R$ be even.
This condition is also sufficient.  To see this, consider the cases where the number of minus signs is $0,$ $2,$ $4.$  
If the number of minuses in $R$ is $0$, then $R$ is an ordinary permutation matrix and $R$ forms part of an automorphism pair.  This follows from the fact that $H'=J-2I,$ where $J$ is the all-ones matrix.  Then $R(J-2I)R^T=J-2I=H',$ and so $(R,R^T)$ is an automorphism.
For the case where the number of minuses is $2,$ observe that the pair $(P,Q)$ with
$$
P=\begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{bmatrix},\qquad Q=\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\end{bmatrix}
$$
produces the same matrix as does the permutation pair $(T,I)$ with
$$
T=\begin{bmatrix} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\end{bmatrix}.
$$
That is, $PH'Q=TH'.$  Since $T$ is its own inverse, $(TP,Q)$ is an automorphism of $H'.$  But if $U$ is any permutation, then $(UTP,QU^T)$ is an automorphism as well.  Since $UTP$ is a permutation matrix with minuses in columns $1$ and $2$, we get an automorphism by taking $R$ to be any monomial matrix with nonzero element $-1$ in columns $1$ and $2$ and nonzero element $1$ in columns $3$ and $4.$  Similarly, by taking the $-1$ elements of $P$ to be in pairs of columns other than $1$ and $2,$ we see that we can take $R$ to be any monomial matrix with elements $-1$ in exactly two columns.
Finally, for the case where the number of minus is $4,$ note that $(-R,-R^T)$ is an automorphism for any permutation matrix $R.$
This establishes that any monomial matrix $R$ with an even number of elements $-1$ gives rise to an automorphism $(R,S).$  The size of the automorphism group is $2^3\cdot4!=192$ since there are $4!$ permutation matrices and $2^3$ sign patterns for columns such that the number of minuses is even.
If you have one pair $(P,Q)$ relating a matrix $H$ to the particular matrix $H'$ above, that is, you have a pair such that $H'=PHQ,$ then you can generate all $192$ such pairs by composing with the automorphisms of $H'.$  If instead, you are interested in relating some other pair of Hadamard matrices, $H_1$ and $H_2,$ you will need pairs $(P_1,Q_1)$ and $(P_2,Q_2)$ such that $H'=P_1H_1Q_1,$ $H'=P_2H_2Q_2.$  Then $H_2=P_2^{-1}RH'SQ_2^{-1}=P_2^{-1}RP_1H_1Q_1SQ_2^{-1}$ for any of the $192$ automorphisms $(R,S)$ of $H'$ described above. In other words, you find all pairs relating $H_1$ to $H_2$ by composing $(P_1,Q_1)$ with $(R,S)$ and then composing the result with $(P_2^{-1},Q_2^{-1}).$
