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)
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 $(P_2^{-1},Q_2^{-1}).$