In category theory there are lots of examples of isomorphisms that cannot be strictified to become identities. For instance, every monoidal category is equivalent to a strict monoidal category, where the associativity and unit isomorphisms are identities, but not every braided monoidal category is equivalent to a strictly braided one where the braiding is an identity.
In higher category theory, constraints are in general (adjoint) equivalences rather than isomorphisms. For instance, the associativity and unit 2-cell constraints for a tricategory (weak 3-category) are equivalences, but the 3-cell constraints (such as the interchanger) are still isomorphisms, since there is "no room" for anything weaker (there being no 4-cells in a tricategory). Every tricategory is equivalent, not to a strict 3-category where all constraints are identities, but to a Gray-category where the associators and unitors are identities but the interchanger is not.
I am looking for an example of a higher-categorical structure containing constraint equivalences which cannot be strictified even to become isomorphisms (not necessarily identities). Since the only nontrivial constraints in a Gray-category are top-dimensional and hence isomorphisms, the first place to look for this would be in some sort of 4-category. But we can also make it more manageable by being somewhat degenerate. A triply degenerate 4-category (exactly one 0-, 1-, and 2-cell) would be (by the delooping hypothesis) a symmetric monoidal category, with no room for any equivalences that aren't isomorphisms, so the next level of complexity seems the first place to look.
A doubly degenerate 4-category should be the same as a braided monoidal bicategory, and by the coherence theorem for tricategories, everyone of those is equivalent to a braided Gray-monoid (a Gray-monoid being a Gray-category with one object). Howver, the braiding in a braided Gray-monoid is, a priori, still only an equivalence, so one way to make this question precise would be:
Is every braided Gray-monoid equivalent to one whose braiding is an isomorphism, rather than merely an equivalence?
Best Answer
I'm surprised that I didn't notice this immediately, and that no one else pointed it out either.
Every category is equivalent to a skeletal one. Therefore, every bicategory is (bi)equivalent to one whose hom-categories are skeletal. But in a bicategory of this sort, every 1-cell equivalence is an isomorphism. Therefore, every (braided, symmetric, ...) monoidal bicategory is monoidally equivalent to one in which all the 1-cell equivalence constraints are isomorphisms.
This is definitely an instance of the "whack-a-mole" aspect of coherence that Peter mentioned, though, since we can't expect to make a bicategory both strict (i.e. a 2-category) and locally skeletal. So it doesn't answer the "one way to make the question precise" that I asked, but it says something about the imprecise version.