Modules with $m \otimes n = n \otimes m$

Let $R$ be a commutative ring. Which $R$-modules $M$ have the property that the symmetry map

$$M \otimes_R M \to M \otimes_R M, ~m \otimes n \mapsto n \otimes m$$

equals the identity? In other words, when do we have $m \otimes n = n \otimes m$ for all $m,n \in M$?

Some basic observations:

1) When $M$ is locally free of rank $d$, then this holds iff $d \leq 1$.

2) When $A$ is a commutative $R$-algebra, considered as an $R$-module, then it satisfies this condition iff $R \to A$ is an epimorphism in the category of commutative rings (see the Seminaire Samuel for the theory of these epis).

3) These modules are closed under the formation of quotients, localizations (over the same ring) and base change: If $M$ over $R$ satisfies the condition, then the same is true for $M \otimes_R S$ over $S$ for every $R$-algebra $S$.

4) An $R$-module $M$ satisfies this condition iff every localization $M_{\mathfrak{p}}$ satisfies this condition as an $R_{\mathfrak{p}}$-module, where $\mathfrak{p} \subseteq R$ is prime. This reduces the whole study to local rings.

5) If $R$ is a local ring with maximal ideal $\mathfrak{m}$ and $M$ satisfies the condition, then $M/\mathfrak{m}M$ satisfies the condition over $R/\mathfrak{m}$ (by 3). Now observation 1 implies that $M/\mathfrak{m}M$ has dimension $\leq 1$ over $R/\mathfrak{m}$, i.e. it is cyclic as an $R$-module. If $M$ was finitely generated, this would mean (by Nakayama) that $M$ is also cyclic. Thus, if $R$ is an arbitrary ring, then a finitely generated $R$-module $M$ satisfies this condition iff every localization of $M$ is cyclic. But there are interesting non-finitely generated examples, too (see 2).

I don't expect a complete classification (this is already indicated by 2)), but I wonder if there is any nice characterization or perhaps even existing literature. It is a quite special property. Also note the following reformulation: Every bilinear map $M \times M \to N$ is symmetric.

The question has an accepted answer at MathOverflow, and perhaps it is time to leave the Unanswered list.