Let $R$ be a normal local domain of dimension $2$ and odd residue characteristic endowed with the action of the finite group $G \cong \mathbb{Z}/2\mathbb{Z}$. Suppose that the ring of invariants $R^G$ has rational singularities. Does this mean that $R$ also have rational singularities?
Of course, one could pose the question more generally where $G$ is finite of order invertible on $R$ (and $R$ is $2$-dimensional). I am not really interested in $R$ of residue characteristic $0$, but if there already are such counterexamples, then they are of course welcome. Also, I recall that if $R$ has rational singularities, then so does $R^G$ (as long as the order of $G$ is invertible in $R$).
My guess is that the answer is 'no', but I don't have a counterexample. A 'yes' would be great, even if one needs to assume additional hypotheses for that.
For context, let me recall that a normal surface singularity like above is rational if some (equivalently, any) desingularization $X \rightarrow \mathrm{Spec}(R)$ with $X$ regular obtained by successively blowing up closed points satisfies $H^1(X, \mathcal{O}_X) = 0$.
