A $C^{*}$ algebra $A$ is graded by $\mathbb{Z}_{n}$ iff it can be acted by $\mathbb{Z}_{n}$. So we associate the $C^{*}$ algebra $A\rtimes \mathbb{Z}_{n}$ to a $\mathbb{Z}_{n}$-graded $C^{*}$ algebra.
Now what about if the grading group is an arbitrary finite group $G$? Is it true to say that: existence of a $G$-graded structure on $A$ is equivalent to existence of a $G$-action on $A$? And is there a natural $C^{*}$ algebra associated to a $G$-graded $C^{*}$ algebra?
What's really happing is this:
Fact 1: The grading group and the operating group are not actually the same, it just looks that way! If you have a finite group $G$ acting on an $k$-algebra $A$ by diagonalisable automorphism, then there is a $\widehat{G}$-grading (where $\widehat{G}:=Hom(G,k^\times)$ with pointwise multiplication) given by $A_\chi := \{a \mid \forall g\in G: ga = \chi(g)a\}$. Conversely: If there is a $\widehat{G}$-grading on $A$, then $G$ acts on $A$ by diagonalisable automorphisms via $ga := \chi(g)a$ for $a\in A_\chi$.
In your case $k=\mathbb{C}$ (in particular: finite groups automatically act diagonalisable) and $\widehat{\mathbb{Z}/n} \cong \mathbb{Z}/n$, but only by a non-canonical isomorphism.
Fact 2: For every $G$ acting on every algebra $A$ one can form the algebra $A\rtimes G$. This can always be done and has not much to do with gradings it seems, but there is a connection: If this group action induces a grading as before, then a ($\widehat{G}$-)graded $A$-module $M$ is the same as a $A\rtimes G$-module on which $G$ acts diagonalisable via $gm := \chi(g)m$ for all $m\in M_\chi$.
I think the finiteness only comes in, when you want $A\rtimes G$ to be a $C^\ast$-algebra.
EDIT: Come to think of it, it might be possible to define a $C^\ast$-algebra $A\rtimes G$ as a twisted tensor product of $A$ with the $C^\ast$-group algebra of $G$. Then the finiteness assumption can be removed.
