Let $M$ denote a finite von Neumann algebra with trace $\tau$, and $L^{2}(M)$ denote the standard (trivial) M-M correspondence (binormal bimodule). The coarse correspondence is $L^{2}(M) \overline{\otimes} L^{2}(M)$ with commuting left and right actions of $M$ given by $x(\xi \otimes \eta)y=(x \xi) \otimes (\eta y)$ for any $x,y \in M$, where $x \xi$ and $\eta y$ refer, of course, to the standard left and right actions on $L^{2}(M)$.

Intuitively, this bimodule is regarded as "coarse" since there is no "connection" between the two copies of $L^{2}(M)$ making it up. If this is the case, what if we consider $L^{2}(M) \overline{\otimes} L^{2}(M) \overline{\otimes} L^{2}(M)$ with elements of $M$ acting only on the left on the left copy and the right on the right copy of $L^{2}(M)$? There still should be no "connection".

Question: As correspondences, is $L^{2}(M) \overline{\otimes} L^{2}(M) \overline{\otimes} L^{2}(M)$ always isomorphic to the coarse correspondence $L^{2}(M) \overline{\otimes} L^{2}(M)$? If so, what is an isomorphism?

As far as I can see, the former is isomorphic to a direct sum of the latter. The intuition suggests that more may be true. (I imagine an answer to this question can be seen by viewing the bimodules as morphisms of M into amplifications of M, but I haven't worked it out.)

EDIT/ADDENDUM: Since this is already pretty silly, why not make it sillier? Is it possible for $M$ to be nonamenable and yet have $L^{2}(M)$ weakly contained in $L^{2}(M) \overline{\otimes} L^{2}(M) \overline{\otimes} L^{2}(M)$?