Let $A$ be a ring. Recall that a module $M$ is called reflexive in case the canonical evaluation map $f_M : M \rightarrow M^{**}$ is an isomorphism. Here $(-)^{*}$ denotes the functor $Hom_A(-,A)$.

In the following I restrict to Artin algebras, but the questions are interesting for more general rings.

Questions:

What are the Artin algebras $A$ such that every simple $A$-module is reflexive?

Is every such algebra selfinjective?

In the answers below you find good evidence that the answer to 2. might be yes.

Being reflexive implies being a 2. syzygy module, so I would guess that 2. is correct.

In https://www.sciencedirect.com/science/article/pii/002186938590198X a positive answer was given in case $A$ has additionally dominant dimension at least two. In fact one can prove that $A$ with dominant dimension at least one and all simple modules reflexive implies that $A$ must be selfinjective: Being reflexive implies being a 2. syzygy module and thus every simple module is a submodule of a projective module, which is equivalent to have dominant dimension at least one for algebras with dominant dimension at least one. But the socle of an indecomposable injective non-projective module has dominant dimension zero. Thus $A$ has to be selfinjective.