= Solution
Put $S=R^G$, the <ring of invariants>. It is an <integral domain> because it is a <subring> of the domain $R$. Embed $\operatorname{Frac}S$ in $\operatorname{Frac}R$ in the evident way.
Suppose $q\in\operatorname{Frac}S$ is an <integral element> over $S$. Its monic equation has coefficients in $S\subseteq R$, so $q$ is integral over $R$. Since $R$ is a <normal domain>, $q\in R$.
Write $q=a/b$ with $a,b\in S$ and $b\ne0$. Each <ring automorphism> in $G$ extends to the <fraction field> by acting on numerator and denominator, and fixes this fraction. Hence $q\in R^G=S$. We have proved that $S$ is integrally closed in its own <fraction field>:
$$
\boxed{R^G\text{ is a normal domain}}.
$$
This <normality of a ring of invariants> argument actually works for any group; finiteness is not needed for this conclusion. For finite $G$, there is additionally an <integral extension> $R^G\subseteq R$, since for each $r\in R$ the <orbit polynomial under a finite automorphism group>
$$
\prod_{g\in G}(T-g(r))
$$
is monic, has coefficients fixed by $G$ and vanishes at $r$. Neither argument divides by $|G|$, so it is valid when the characteristic divides the group order. Here normality means integral closedness of a domain.
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