280: my thought process: consider a monoid where morphisms I->I are elements and multiplication is composition. Then the exercise is asking to show that this monoid is commutative. This sounds like the Eckmann-Hilton argument.
Concretely (but with mistakes I need help with) note that \\((f \otimes g) \circ (a \otimes b) = (f \circ a) \otimes (g \circ b) \\) which is to say that \\( \otimes \\) is a functor as Anindya noted (but I suppose this also needs to be shown). I've described this as an equality, but is that too strong?
Then \\(id \\) = \\(id \circ id\\) = \\( (I \otimes id) \circ (id \otimes I) \\) = \\( (I \circ id) \otimes (id \circ I) \\) = \\( I \otimes I \\) = \\(I \\) . Again, lots of these equalities are clearly wrong. Not least because there's no morphism from X to \\( I \otimes X \\). Moreover, id is a morphism and I is an object. They're definitely not equal, although for id restricted to I, they might be isomorphic.
One continues as on https://en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument to show that a.b = b.a: I'm not bothering with that yet, because I want to debug my conceptual confusion this far / check if I'm doing something reasonable.
Edit 1: hmm okay, \\( (id \circ I) \\) isn't even well-typed.
Edit 2: aha! maybe what I should show is, instead: for any morphism f : \\(I \to I \\), f is isomorphic to \\( id \\). Then commutativity of fg and gf follows trivially. This is easier: \\( f \approx (f \otimes f) = ((f \circ id) \otimes (id \circ f) ) = ((f \otimes id) \circ (id \otimes f) ) = (id \otimes id ) \approx id \\) . This is still clearly wrong somehow, but maybe less wrong?