Two morphisms of category $\ \mathbf C\ $ are isomorphic to one
another $\ \Leftarrow:\Rightarrow\ $ they are the opposite edges that are drawn horizontally (aimed East) of a commutative square that has the vertical edges (aimed North) being isomorphisms of $\ \mathbf C$.


**Problem** What is the minimum total number of morphisms
of a category such that there are isomorphic morphisms
$\ f\ $ and $\ u,\ $ and another isomorphic pair $\ g\ $ and $\ v,\ $ and the compositions $\ g\circ f\ $ and
$\ v\circ u\ $ exist but are not isomorphic?


I have an example of a category, as described above, that has a total number of $27$ morphisms. (*No, I've touched NO computer* :) ).




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AN ADDITIONAL NOTE:


Now, that @HenrikRüping has provided his excellent [example](https://mathoverflow.net/a/404994) (most likely minimal), let me mention that my example was a monoid too (but of course) of all maps into itself of a 3-element set.



Isomorphic morphisms. A 27-morphism category
I see one example with 7 morphisms. It is a subcategory of the category of groups. The only object is the the Klein 4-group $(\mathbb{Z}/2)^2$, and the morphisms are generated by the two projections and the flip. That monoid has 7 elements, and the two projections are conjugate, and hence they are isomorphic. However $pr\_1\circ pr\_1=pr\_1$ is not isomorphic to $pr\_1\circ pr\_2=0$.


Edit: I miscounted the number of elements. Originally I thought of it as a semidirect product (which it is not). However in terms of matrices, it should consist exactly of the elements
$\pmatrix{1&0\\0&1}$,$\pmatrix{1&0\\0&0}$,$\pmatrix{0&0\\0&1}$,$\pmatrix{0&0\\0&0}$,$\pmatrix{0&1\\1&0}$,$\pmatrix{0&0\\1&0}$,$\pmatrix{0&1\\0&0}$.


