The strategy of the proof is to use \'etale localization to reduce to the case where $R \to R^I$ is a local isomorphism at $R^I \cap \mathfrak p$. Let $R^G \to A$ be an \'etale ring map. We claim that if the result holds for the action of $G$ on $A \otimes_{R^G} R$ and some prime $\mathfrak q'$ of $A \otimes_{R^G} R$ lying over $\mathfrak q$, then the result is true. \medskip\noindent To check this, note that since $R^G \to A$ is flat we have $A = (A \otimes_{R^G} R)^G$, see More on Algebra, Lemma \ref{more-algebra-lemma-base-change-invariants}. By Lemma \ref{lemma-inertia-base-change} the group $I$ does not change. Then a second application of More on Algebra, Lemma \ref{more-algebra-lemma-base-change-invariants} shows that $A \otimes_{R^G} R^I = (A \otimes_{R^G} R)^I$ (because $R^I \to A \otimes_{R^G} R^I$ is flat). Thus $$ \xymatrix{ \Spec((A \otimes_{R^G} R)^I) \ar[d] \ar[r] & \Spec(R^I) \ar[d] \\ \Spec(A) \ar[r] & \Spec(R^G) } $$ is cartesian and the horizontal arrows are \'etale. Thus if the left vertical arrow is \'etale in some open neighbourhood $W$ of $(A \otimes_{R^G} R)^I \cap \mathfrak q'$, then the right vertical arrow is \'etale at the points of the (open) image of $W$ in $\Spec(R^I)$, see Descent, Lemma \ref{descent-lemma-smooth-permanence}. In particular the morphism $\Spec(R^I) \to \Spec(R^G)$ is \'etale at $R^I \cap \mathfrak q$. \medskip\noindent Let $\mathfrak p = R^G \cap \mathfrak q$. By More on Algebra, Lemma \ref{more-algebra-lemma-one-orbit} the fibre of $\Spec(R) \to \Spec(R^G)$ over $\mathfrak p$ is finite. Moreover the residue field extensions at these points are algebraic, normal, with finite automorphism groups by More on Algebra, Lemma \ref{more-algebra-lemma-one-orbit-geometric}. Thus we may apply More on Morphisms, Lemma \ref{more-morphisms-lemma-etale-makes-integral-split} to the integral ring map $R^G \to R$ and the prime $\mathfrak p$. Combined with the claim above we reduce to the case where $R = A_1 \times \ldots \times A_n$ with each $A_i$ having a single prime $\mathfrak q_i$ lying over $\mathfrak p$ such that the residue field extensions $\kappa(\mathfrak q_i)/\kappa(\mathfrak p)$ are purely inseparable. Of course $\mathfrak q$ is one of these primes, say $\mathfrak q = \mathfrak q_1$. \medskip\noindent It may not be the case that $G$ permutes the factors $A_i$ (this would be true if the spectrum of $A_i$ were connected, for example if $R^G$ was local). This we can fix as follows; we suggest the reader think this through for themselves, perhaps using idempotents instead of topology. Recall that the product decomposition gives a corresponding disjoint union decomposition of $\Spec(R)$ by open and closed subsets $U_i$. Since $G$ is finite, we can refine this covering by a finite disjoint union decomposition $\Spec(R) = \coprod_{j \in J} W_j$ by open and closed subsets $W_j$, such that for all $j \in J$ there exists a $j' \in J$ with $\sigma(W_j) = W_{j'}$. The union of the $W_j$ not meeting $\{\mathfrak q_1, \ldots, \mathfrak q_n\}$ is a closed subset not meeting the fibre over $\mathfrak p$ hence maps to a closed subset of $\Spec(R^G)$ not meeting $\mathfrak p$ as $\Spec(R) \to \Spec(R^G)$ is closed. Hence after replacing $R^G$ by a principal localization (permissible by the claim) we may assume each $W_j$ meets one of the points $\mathfrak q_i$. Then we set $U_i = W_j$ if $\mathfrak q_i \in W_j$. The corresponding product decomposition $R = A_1 \times \ldots \times A_n$ is one where $G$ permutes the factors $A_i$. \medskip\noindent Thus we may assume we have a product decomposition $R = A_1 \times \ldots \times A_n$ compatible with $G$-action, where each $A_i$ has a single prime $\mathfrak q_i$ lying over $\mathfrak p$ and the field extensions $\kappa(\mathfrak q_i)/\kappa(\mathfrak p)$ are purely inseparable. Write $A' = A_2 \times \ldots \times A_n$ so that $$ R = A_1 \times A' $$ Since $\mathfrak q = \mathfrak q_1$ we find that every $\sigma \in I$ preserves the product decomposition above. Hence $$ R^I = (A_1)^I \times (A')^I $$ Observe that $I = D = \{\sigma \in G \mid \sigma(\mathfrak q) = \mathfrak q\}$ because $\kappa(\mathfrak q)/\kappa(\mathfrak p)$ is purely inseparable. Since the action of $G$ on primes over $\mathfrak p$ is transitive (More on Algebra, Lemma \ref{more-algebra-lemma-one-orbit}) we conclude that, the index of $I$ in $G$ is $n$ and we can write $G = eI \amalg \sigma_2I \amalg \ldots \amalg \sigma_nI$ so that $A_i = \sigma_i(A_1)$ for $i = 2, \ldots, n$. It follows that $$ R^G = (A_1)^I. $$ Thus the map $R^G \to R^I$ is \'etale at $R^I \cap \mathfrak q$ and the proof is complete.