Coherence Conditions and Strict Monoidal Equivalences Consider the following: two monoidal categories $({\cal C},\otimes)$, and $({\cal D},\odot)$, and a functor $F:{\cal C} \to {\cal D}$, that gives an equivalence (of ordinary categories) between ${\cal C}$ and ${\cal D}$. Moreover, let us assume that we also have 
$$
F(X \otimes Y) = F(X) \odot F(Y),  ~~~~~ \text{ for } X,Y \in {\cal C}.
$$
Does it automatically follow from this that $F$ is a (strict) equivalence of mondoidal categories? That is to say, do the necessary coherence conditions hold automatically?
To me this seems obviously true. However, there have times when what were "obvious truths" to me in general category theory, turned out to be anything but. So, I'd like a voice of confirmation. Thanks!
 A: I haven't been able to come up with an explicit counterexample, but I believe one exists. The problem amounts to finding a category $\mathcal{C}$ equipped with a tensor product $\otimes$ and a unit $I$ such that there are two distinct coherence structures $(\lambda, \rho, \alpha)$. We make $\mathcal{C}$ into a monoidal category using the first coherence structure, and we take $\mathcal{D}$ to be same as $\mathcal{C}$ but with the second coherence structure $(\lambda', \rho', \alpha')$ instead. Suppose $\alpha \ne \alpha'$. Then, the "identity" functor $F : \mathcal{C} \to \mathcal{D}$ is an isomorphism of categories but fails to be a strong monoidal functor under the obvious identity transformations $I \to F I$, $F A \otimes F B \to F (A \otimes B)$ because the diagram below fails to commute:
            
(A similar argument can be made for the case $\lambda \ne \lambda'$ or $\rho \ne \rho'$.) 

Basically, what I would like to say is that because the theory of monoidal categories with two coherence structures is a consistent essentially algebraic theory, such a category $\mathcal{C}$ exists: simply take the free such category generated by one object. Unfortunately, one still has to prove that this category has two distinct coherence structures!
