Assume we have a noncommutative ring $R$ with exactly 2 non-isomorphic simple left modules $S\_1$ and $S\_2$ (up to isomorphism) and an $R$-bimodule $M$, which switches the simples, i.e. $M\otimes\_R S\_1=S\_2$ and $M\otimes\_R S\_2=S\_1$.


Then we have $Hom\_R(S\_i,M\otimes\_R S\_i)=0$ by Schur's lemma ($\*$).


Now assume $F$ is an arbitrary left $R$-module of finite length. What can be said about $Hom\_R(F,M\otimes\_R F)$ with the help of ($\*$)?


If $F$ is a quotient of $R$, i.e. we have a surjection $R\rightarrow F \rightarrow 0$, then we get an injection $0\rightarrow Hom\_R(F,M\otimes\_R F)\rightarrow Hom\_R(R,M\otimes\_R F)$. Is there a method, using ($\*$) to decide whether this is even an isomorphism? 


So for example, if $F$ is one of the simples or a direct sum of one of the simples, then $Hom\_R(F,M\otimes\_R F)=0$ so the map is not an isomorphism. Otherwise both simple modules occur at least once in a Jordan-Hölder composition series. What can be said in this case? 


If this question is too broad, the ring $R$ I'm interested in is the following subring of $M\_2(A)$:
\begin{pmatrix}A &A \\ xA &A \end{pmatrix}
where $A$ is a complete regular local ring of dimension 2, and $x\in A$ s.t. $A=\mathbb{C}[[x,y]]$. The $R$-bimodule $M$ is given as the following $R$-submodule in $M\_2(Quot(A))$:
\begin{pmatrix}A &x^{-1}A \\ A &A \end{pmatrix}



Are morphisms of finite length modules determined by the behaviour of the simple modules?
The question is too broad in general, but I believe that there is a nice answer for your particular example. The first thing to note is that the given $M$ is a twisted bimodule. Namely, let $\sigma$ be the automorphism of $R$ given by $$\sigma : \begin{pmatrix} r & s \\ xt & u \end{pmatrix} \mapsto \begin{pmatrix} u &t \\ xs &r \end{pmatrix}.$$
 Then we can define a bimodule isomorphism $\varphi : {}\_1 R\_{\sigma} \rightarrow M$ by
 $$\varphi : \begin{pmatrix} a &b \\ c &d \end{pmatrix} \mapsto \begin{pmatrix} b & x^{-1}a \\ d & c \end{pmatrix}.$$


It follows that tensoring with $M$ can be thought of as the endofunctor on the category of (left) R-modules that is induced by the automorphism $\sigma^{-1}$ ($= \sigma$). Consequently, the modules $F$ and $M \otimes\_R F$ have the same structure (in a sense), but with the opposite composition factors.
(To contrast with the general situation, the assumption that tensoring with $M$ swaps the simples, alone, does not guarantee that tensoring with $M$ preserves the length of finite-length modules.)


Thus, if you know the structure of $F$, it should not be hard to see if there are any nonzero morphisms from $F$ to $M \otimes\_R F$. For example, if $F$ contains a factor of $S\_1$ in its top and a factor of $S\_2$ in its socle, then $M \otimes\_R F$ contains a factor of $S\_1$ in its socle, and there will be a nonzero map from $F$ to $M \otimes\_R F$. On the other hand, if $F$ is uniserial with composition factors $S\_1, S\_2, S\_1, S\_1$ (from the top down), then $\mbox{Hom}\_R(F,M\otimes\_R F) = 0$.


However, for a finite-length module $F$ over the given $R$, it appears plausible that $\mbox{Hom}\_R(F, M \otimes\_R F)=0$ if and only if all composition factors of $F$ are isomorphic. I do not have a proof at the moment, but one might try to show that for any $F$ of finite length with $S\_1$ in its top, at least one of the three cases occurs: 1) all composition factors of $F$ are isomorphic to $S\_1$; 2) the socle of $F$ contains a copy of $S\_2$; or 3) the length-$2$ uniserial module with composition factors $S\_1, S\_2$ is a quotient of $F$ and the length-$2$ uniserial with composition factors $S\_2, S\_1$ is a submodule of $F$. In cases 2) and 3) there will always be a nonzero map from $F$ to $M \otimes\_R F$. (Note: this suggestion is based on my belief that $R$ is isomorphic to the completed path algebra of the quiver with 2 vertices, a loop at each vertex and a pair of arrows connecting the vertices in each direction, modulo the relations that the two paths of length 2 from each vertex to the other are equal. If this is incorrect, then the structure of the finite-length $R$-modules may be more complicated than I envision.)


We can also see that for a cyclic module $F$, in order for the map $\mbox{Hom}\_R(F,M\otimes\_R F) \rightarrow \mbox{Hom}\_R(R,M \otimes\_R F)$ to be an isomorphism, there must be an epimorphism $F \rightarrow M \otimes\_R F$, which would have to be an isomorphism since these two modules have the same length.


