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EDIT: Inspired by Charles's answer, here's a closely related question. I'm really interested in invertible objects in the derived category, and in the derived category dualizable objects can be represented by finite chain complexes of finite-rank projective modules. Over a local ring, then, all dualizable objects have an Euler characteristic which is an integer (as opposed to an arbitrary element of $R$). Since as Charles noted, the braid map of an invertible object can be identified with its Euler characteristic as a dualizable object, this implies that the braid map of any invertible object in the derived category of a local ring is $\pm 1$ (and so if you're willing to suspend your objects if necessary, you can assume it is 1).

Thus I would be satisfied with a conceptual answer to the following question: why is the Euler characteristic of a dualizable object in the derived category of a local ring always an integer? (It may be more natural to not assume that the ring is local, in which case you should replace "integer" with "integral linear combination of idempotents".)
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Why is the symmetric monoidal structure on invertible modules strict?

Let $N$ be an object in a symmetric monoidal category. Then the braid map $N\otimes N\to N\otimes N$ is almost never the identity, and this is the obstruction to making a symmetric monoidal category into a "strict" symmetric monoidal category, in which the functor $\otimes$ is commutative on the nose. For example, when $\otimes$ is just the categorical product in the category of sets (or any other concrete category), this is the map $(x,y)\mapsto(y,x)$ which is almost never the identity.

However, one case in which the braid map is the identity is in the category of invertible modules over a commutative ring $R$, as a full subcategory of all modules. Indeed, the braid map $R\otimes R\to R\otimes R$ is the identity, and any invertible module $I$ is locally isomorphic to $R$, so the braid map $I\otimes I\to I\otimes I$ is locally equal to the identity and hence equal to the identity.

What I'd like to have is a more conceptual explanation for why the braid map is the identity for invertible modules, which does not use the fact that they are locally free (indeed, I'm interested in this because I want to use this to prove invertible modules are locally free in a more general setting). Unfortunately, the proof cannot just be abstract nonsense involving invertibility--for example, if we work with graded modules over a commutative ring instead of ordinary modules and use the usual sign conventions, then the braid map will be -1 rather than 1 on invertible modules concentrated in odd degrees. Does anyone know of a better explanation, or know a reason I shouldn't expect there to be one?