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Is recognizing if a Latin square is isotopic to its transpose more efficient than computing its symmetry group?

Ihrig and Ihrig (2007) described a mathematical method for determining if a Latin square is isotopic to its transpose (where isotopic Latin squares vary by permuting the rows, columns and symbols). They don't go into the complexity of this method.

I'm not really familiar with Ihrig$^2$'s work, but if I wanted to know if a Latin square is isotopic to its transpose, I'd compute its autoparatopism group using the McKay, Meynert, Myrvold (2006) method (which computes the automorphism group of a corresponding graph using Nauty) and look at the list of autoparatopisms one by one. This is basically the default way to do it.

Question: Is recognizing if a Latin square is isotopic to its transpose more efficient than computing its symmetry group?

I'm wondering if Ihrig$^2$'s method, or any other known method, is more efficient than the default way.