Given a Cohen-Macaulay ring $R$ over a field of characteristic zero and $G$
a reductive algebraic group acting on $R$, then the ring of ivanriants $R^G$
is also Cohen-Macaulay. This is known as Hochster-Roberts Theorem.
It is also known that if $X=Spec(R)$, has rational singularities, then 
$Y=Spec(R^G)$ has rational singularities too. 


I am trying to find a reciprocal of this Theorem.
I mean, If we have a reductive algebraic group $G$ acting on a ring $R$
over a field of characteristic zero and the ring of invariants $R^G$ is Cohen-Macaulay,


1.- Is $R$ Cohen-Macaulay? 


2.- Do exists known conditions over $R$ which ensures the above implication to hold?


I am particularly interested in the case when $G$ is a finite group.


Thanks in advance.



Hochster-Roberts Theorem reciprocal
This is not true in general. 


Indeed, there exists $\mathbb{Q}$-Gorenstein rings $S$ with rational singularities such that their canonical covers do not have rational singularities, see for instance *Cyclic covers of rings with rational singularities.* by Anurag K. Singh. Generally speaking taking Veronese's of graded rings can remove problematic local cohomology that stops rings from being Cohen-Macaulay.


On the other hand, if $S \subseteq R$ is a finite extension of normal rings in characteristic zero. which is etale in codimension $1$ and such that $S$ is log terminal, then $R$ is also log terminal (and hence Cohen-Macaulay). This is sort of the only statement in this direction that I know. Of course, requiring $S$ to be log terminal is a lot stronger than simply requiring Cohen-Macaulay, but maybe it holds in your case.


