Oho. Recall the symmetries I mentioned in #30 -- these are just automorphisms on the given preorder.

Say we have a monoidal preorder \\(\langle P, \le, \otimes, 1 \rangle\\) and an automorphism \\(A\\) on the underlying preorder \\(P'\\) of \\(P\\), and define \\(x \otimes_A y\\) by \\(A(A^{-1}(x) \otimes A^{-1}(y))\\). Then \\(\otimes_A\\) also makes \\(P'\\) a monoidal preorder. It is clear that \\(A(1)\\) is the new unit, so we just have to show associativity.

Consider \\((x \otimes_A y) \otimes_A z\\). This expands to the wild mess of \\(A(A^{-1}(A(A^{-1}(x) \otimes A^{-1}(y))) \otimes A^{-1}(z))\\), which simplifies to \\(A(A^{-1}(x)\otimes A^{-1}(y) \otimes A^{-1}(z))\\). Similarly for \\(x \otimes_A (y \otimes_A z)\\). Thus, \\(\otimes_A\\) is associative.

This is awesome: for every _automorphism_ on the underlying preorder we get an _isomorphism_ between monoidal preorders. This also helps to formalize my hunch from #30: the symmetries of the preorder _do_ help to constrain the possible monoidal structures, because if you can find _one_ monoid, then every of the monoids induced by the automorphisms must also be valid for the preorder.

It seems just possible that an automorphism of preorders may also give an automorphism of monoidal preorders, meaning that particular automorphism might not actually give a useful constraint, but I haven't been able to show or deny this. I also haven't considered whether mere isomorphisms between preorders give the same construction; it would be nice, since it would let us use the fact that the bowtie is isomorphic to its dual.