If two monoidal categories are monoidally isomorphic then if one is strict so is the other. I am trying to show this fact as it seems to be assumed in Remark 2.8.6 of the tensor categories book. Any help would be appreciated.
Recall that a category is strict if the associator is the identity morphism so $X \otimes Y \otimes Z $ is well defined.
By monoidally Isomorphic I mean that there are monoidal functors $F:\mathcal{C}\rightarrow\mathcal{D}$ and $G:\mathcal{D}\rightarrow\mathcal{C}$ such that $FG=1$ and $GF=1$.
So far all I could show is that if $J_{X,Y}:F(X) \otimes F(Y)\rightarrow F(X \otimes Y)$ and $K_{X,Y}$ are the relevant natural isomorphisms making $F$ and $G$ monoidal functors then $F(K_{X,Y})=J_{G(X),G(Y)}$ and $G(J_{X,Y})=K_{F(X),F(Y)}$.
 A: This isn't true as stated. Consider the category $\mathcal{2} = \{0, 1\}$ whose two objects are declared to be isomorphic by a unique isomorphism (so there are a total of 4 morphisms in the category). That makes $\mathcal{2}$ a thin category, so every diagram commutes, which will make verifying everything much easier.
We can then put two different monoidal structures on $\mathcal{2}$. We'll take $\mathcal{C}$ to be $\mathcal{2}$ with $x \otimes y$ as simply multiplication (of $0$ and $1$). This is clearly a strict monoidal category because multiplication has a strict unit and strict associativity.
On the other hand, take $\mathcal{D}$ to be $\mathcal{2}$ with the monoidal product $x \odot y := 0$. This is again strictly associative, but it doesn't have a strict unit: $x \odot 1 = 0 \not = 1$. It still has a unit though. Either $0$ or $1$ will work since $x \odot y = 0 \simeq x$ regardless of what $x$ and $y$ are. This isomorphism is natural and all the necessary diagrams commute because $\mathcal{2}$ is a thin category.
Now take $F: \mathcal{C} \to \mathcal{D}$ to be the identity on $\mathcal{2}$. It's a monoidal functor since all the necessary morphisms exist and all the necessary diagrams commute (because $\mathcal{2}$ is thin and any two objects have a morphism between them). Similarly define $G: \mathcal{D} \to \mathcal{C}$ to also be the identity (on $\mathcal{2}$).
$FG = 1$ and $GF = 1$ since both are the identity. Thus, $\mathcal{C}$ and $\mathcal{D}$ are monoidally isomorphic (in the given sense). But $\mathcal{C}$ has a strict monoidal structure, while $\mathcal{D}$ does not.

I'm of the opinion that the definition of monoidally isomorphic given in Tensor Categories is too weak to be useful. It should also require that the functors in question are strict monoidal functors. That means that they take the unit to the corresponding unit and the product of two objects to the corresponding product strictly. Being an isomorphism is already a "strict" condition, so we really ought to make that strictness also respect the monoidal structure.
If you change the definition like that, it becomes almost obvious that if two categories are monoidally isomorphic and one is strictly mondoial, then so is the other.

I'll finish off by noting that strengthening the definition of monoidally isomorphic only helps the point in Remark 2.8.6. Now it's even harder for two categories to be monoidally isomorphic, so proving that the category presented there isn't monoidally isomorphic to a strict mondoidal category carries through with the stronger condition too.
