Correspondence of grading and group actions Let $k$ be a field of characteristic zero with $n$-th roots of unity, $R$ a $k$-algebra and $G := \mathbb{Z}/n\mathbb{Z}$.
To give an action of $G$ on $R$ is equivalent to grading $R$ by $G$. (To see this, look at the eigenspaces $R_\gamma := \{r \in R \mid \forall g \in G: g(r) = \gamma(g)r \}$ for $\gamma \in \hat{G} \cong G$ which gives a $G$-grading on $R$. Conversely, we let $g$ act on $R_\gamma$ by multiplication by $\gamma(g)$.)
In other words there is the following correspondence:
$$\{G\text{-action on } R\} \leftrightarrow \{G\text{-grading on } R\}$$
Now let $H$ be an arbitrary finite group and $\theta : H \to \operatorname{Aut}(G)$.
I am interested in a correspondence of the following type:
$$\{G \rtimes_\theta H \text{ action on } R\} \leftrightarrow \{H \text{-action on } \bigoplus_{\gamma \in G} R_\gamma \text{ with} \textit{ additional properties?}\}$$
There are obviously some additional properties missing on the right hand side because there has to be some kind of compatibility between the $H$-action and the grading. These additional properties probably correspond to the map $\theta$ on the left hand side but I fail at working them out.
(I expect them to be along the lines of „$H$ does not change the grading“ or similar.)
Any help would be greatly appreciated.
 A: It is really important that you distinguish the two copies of $\mathbb{Z}/n\mathbb{Z}$ appearing here: one is the group of characters, or said another way the Pontryagin dual, of the other, and there is no natural isomorphism between them. 
This is important for understanding what is happening in your question. The action of $H$ on $G$ induces a dual action of $H$ on the Pontryagin dual $\hat{G}$, which is the actual group $R$ is being graded by, and the additional condition that $H$ has to satisfy is a compatibility with this action, namely that the action of $H$ on a homogeneous element in $R_{\gamma}, \gamma \in \hat{G}$ lands in $R_{h \gamma}$ (again, this is the dual action, not the original action). 
This is also important for understanding what happens more generally. You can talk about a $\mathbb{Z}/n\mathbb{Z}$ grading over a field not necessarily containing all roots of unity; what is the analogue on the action side? It turns out to be an action of the group scheme $\mu_n$ of $n^{th}$ roots of unity, which is now the Cartier dual of $\mathbb{Z}/n\mathbb{Z}$, and now the two are genuinely non-isomorphic as group schemes. 
