Finding intercalates within a (reduced) latin square I need confirmation on if my intuition on finding intercalates is correct, suppose we have the following reduced latin square
\begin{bmatrix}
    1  & 2 & 3 & 4 \\
    2       & 4 & 1 & 3  \\
    3       & 1 & 4 & 2\\
    4       & 3 & 2 & 1 
\end{bmatrix}
we know from a theorem that for a latin square of order n(even), there is at most $\frac{n^2(n-1)}{4}$ intercalates; in this case 12.
I am not sure if two 2 x 2 sub matrixes can be considered the same intercalate if they are flipped and rotated versions of each other. From the square above we get: (e.g R1 = row 1)
R1,R2
\begin{bmatrix}
    1  & 2 & |  3&4\\
    2       & 4 & |1&3 
\end{bmatrix}
R1,R3
\begin{bmatrix}
    1  & 2 & |  3&4\\
    3       & 1 & |4&2 
\end{bmatrix}
R1,R4\begin{bmatrix}
    1  & 2 & |  3&4\\
    4       & 3 & |2&1 
\end{bmatrix}
R2,R3
\begin{bmatrix}
    2  & 4 & |  1&3\\
    3      & 1 & |4&2 
\end{bmatrix}
R2,R4
\begin{bmatrix}
    2  & 4 & |  1&3\\
    4      & 3 & |2&1 
\end{bmatrix}
R3,R4
\begin{bmatrix}
    3  & 1 & |  4&2\\
    4       & 3 & |2&1 
\end{bmatrix}
And only keeping ones that are not flips and rotations of each other (I keep 1 copy). These should be the only intercalates right? So 6 of them.
\begin{bmatrix}
    1&2&|3&4&|&3&4&|&1&2&|1&2&|&2&4&\\
    2&4&|1&3&|&4&2&|&3&1&|4&3&|&3&1&
\end{bmatrix}

None of these are intercalates. As defined in this paper (for example), we're looking for two rows $i,j$ and two columns $x,y$ such that $L_{i,x} = L_{j,y}$ and $L_{j,x} = L_{i,y}$. There are four of these in your example:


*

*$\{i,j\} =\{1,4\}$ and $\{x,y\} = \{1,4\}$: $$\begin{bmatrix}1 & 4 \\ 4 & 1\end{bmatrix}$$

*$\{i,j\} =\{2,3\}$ and $\{x,y\} = \{2,3\}$: $$\begin{bmatrix}4 & 1 \\ 1 & 4\end{bmatrix}$$

*$\{i,j\} =\{1,4\}$ and $\{x,y\} = \{2,3\}$: $$\begin{bmatrix}2 & 3 \\ 3 & 2\end{bmatrix}$$

*$\{i,j\} =\{2,3\}$ and $\{x,y\} = \{1,4\}$: $$\begin{bmatrix}2 & 3 \\ 3 & 2\end{bmatrix}$$
