Studying diophantine magic squares

Question

A bit of context

I'm analyzing a type of cryptarithmetic puzzle that consist in the simultaneous resolution of 6 cryptarithms arranged in a magic square fashion.
Here's an example:

Solution:

7008, 612, 6396 // 584, 571, 13 // 12, 41, 492

My goal is possibly finding how many puzzles are there (with numbers up to $k$ digits) net of symmetries.
I've started calculating how many unique structures of operations are there up to symmetries.

Since (assuming the number $0$ is never used otherwise the puzzle would be trivialized) the result of each of the four operations can always be swapped with the first operand and the operation inverted

• you can always swap the first column(row) with the last one and invert all the row(column) operations.

  These four puzzles are equivalent
  
    +   =       -   =       *   =       /   =  
  /   -   +   +   -   /   *   +   -   -   +   *
    -   =       +   =       -   =       +   =  
  =   =   =   =   =   =   =   =   =   =   =   =
    *   =       /   =       +   =       -   =    
  

In addition

• if a column(row) consist of only $+$ and $\times$ you can swap the first column(row) with the second
• if a column(row) consist of only $-$ and $\div$ you can swap the second column(row) with the third

And of course

• you can always transpose the entire matrix

I've written a code that given a structure finds all the members of its equivalence class.
If my analysis and my code are correct, there are $433$ unique structures of operations.

Hence $433$ different systems of diophantine equations. ($9$ of them consist of only $+$ and $-$ and $9$ of them consist of only $\times$ and $\div$, those are trivial)

The question

A general form of this class of diophantine equations is:

$$ (a \star_1 b) \star_2 (c \star_3 d) = (a \star_4 c) \star_5 (b \star_6 d) $$

$a,\ b,\ c,\ d,\ (a \star_1 b),\ (c \star_3 d),\ (a \star_4 c),\ (b \star_6 d),\ (a \star_1 b) \star_2 (c \star_3 d) \in \mathbb{Z}^+$

How can I find the solutions or determine there are none? Is there a simple algorithm?