Let $A$ be a local symmetric finite dimensional algebra and $M$ an $A$-module with at least two non-isomorphic indecomposable non-projective summands.
In case $\Omega^1(M) \cong M$ in the stable category of $A$, we have that $\underline{End_A}(M)$ is selfinjective by the Auslander-Reiten forumla.
Can we give a converse to this statement? At least under some possibly stronger assumptions on $A$ and $M$.
So the general question might be stated as follows:
Question 1: Can we give a module theoretic characterisation on those $M$ such that $\underline{End_A}(M)$ is selfinjective (probably we should add also connected and non-semisimple to the conditions)?
Is there an example such that $\underline{End_A}(M)$ is selfinjective but we do not have $\Omega^1(M) \cong M$?
(The assumptions that $A$ is local symmetric might be relaxed to A being just symmetric, I just looked at local algebras. Maybe if we drop the local condition we should assume that the stable endomorphism ring is connected and non-semisimple.)
Example: Let $A=K[x]/(x^n)$. Then we really should have an equivalence. Namely such $M$ has $\Omega^1(M) \cong M$ in the stable category of $A$ if and only if $\underline{End_A}(M)$ is selfinjective. In case we take for M the direct sum of all $A$-modules we get the preprojective algebra of Dynkin type $\mathcal{A}$ here.
Question 2: Is there an easy proof of: $M$ has $\Omega^1(M) \cong M$ in the stable category of $A$ if and only if $\underline{End_A}(M)$ is selfinjective for $A=K[x]/(x^n)$ and $M$ having at least two indecomposable non-projective summands? Easy means a proof that possibly avoids the calculation of the endomorphism rings $\underline{End_A}(M)$ by quiver and relations and uses some sort of theoretical argument.
