Let $G$ be a finite group acting on a commutative ring $R$ via ring maps. In doubt, one can assume $R$ to be noetherian or regular if one wants. Let $P$ be a $1$-dimensional free $R$-module with a $G$-action satisfying $g(r\cdot x) = g(r)\cdot g(x)$ for $r\in R$ and $x\in P$, i.e. what some people call a $R$-semilinear $G$-action. Examples include the following, where always $G= \mathbb{Z}/2 = \langle t\rangle$:

1) $R= \mathbb{Z}[i]$ with $t(i) = -i$. We act on $P \cong \mathbb{Z}[i]$ via interchanging $1$ and $i$. 

2) $R = \mathbb{C}[[X,Y]]$, $t(X) = Y$, $t(Y) = X$. We act on some power series $f\in P \cong \mathbb{C}[[X,Y]]$ via $t(f) = f^{op} \cdot e^{X-Y}$, where $f^{op}$ denotes interchanging $X$ and $Y$. 

3) $R = k[X^{\pm 1}]$ for any field $k$ and $t(X) = X^{-1}$. We act on $P \cong k[X^{\pm 1}]$ by $t(1) = X$ (or, more generally, $t(X^n) = X^{-n+1}$). 

These representations are quite different, but they have all in common that a tensor power of them is "trivial", i.e. isomorphic to $R$ as an $R$-module with semilinear $G$-action. In the first example, $P^{\otimes_R 2}$ (with the diagonal action) is trivial in this sense as $t(i\otimes 1) = 1 \otimes i = i\otimes 1$ and thus $1\mapsto i$ is a $G$-equivariant isomorphism $R\to P$. The second example is even trivial itself as $1+e^{X-Y}$ is an invariant generator of $P$. In the third example, $X(\otimes 1)$ is an invariant element of $P\otimes_R P$.

This suggests the following question:

>**Question:**  Is there always some tensor power $P^{\otimes_R n}$ (with diagonal action) which is "trivial", i.e. isomorphic to $R$ as an $R$-module with semilinear $G$-action? Can $n$ be chosen to be $|G|$?

This is certainly true in the special case that *$G$ acts trivially on $R$*: Then $G$ acts $R$-linearly on $P$, i.e. each element $g\in G$ acts via multiplication by an element $r_g$. As $g^{|G|} = e_G$, all elements $r_g$ are $|G|$-th roots of unity in $R$. Thus, $P^{\otimes_R |G|}$ has trivial $G$-action.

In general, one might also ask for something slightly weaker, which might be more sensible if $Pic(R^G)$ has infinite order:

>**Question:** Is there some tensor power $P^{\otimes_R n}$ which is induced up from an invertible module over $R^G$? More precisely, one might ask that $(P^{\otimes_R n})^G$ is an invertible $R^G$-module and $R\otimes_{R^G}(P^{\otimes_R n})^G \to (P^{\otimes_R n})$ an isomorphism. 

This seems to be true in the case that *the order of $G$ is prime to all residue characteristics of $R$*: An $1$-dimensional free $R$-module $P$ with semilinear $G$-action is equivalent to a line bundle $L$ on the stack quotient $Spec R//G$. The coarse moduli space of $Spec R//G$ is $\pi: Spec R//G \to Spec R^G$. Thus, we have to show in this case that $\pi_*L^{\otimes n}$ is a line bundle on $Spec R^G$ and $\pi^*\pi_*L^{\otimes n} \to L^{\otimes n}$ is an isomorphism for some $n$. This is shown in Fulton-Olsson's article [The Picard Group of $\mathcal{M}_{1,1}$][1] Lemma 2.2 if the stabilizer of each geometric point $x$ of $Spec R//G$ acts trivially on $L_x^{\otimes}$. This is always true for $n= |G|$.

I am not too optimistic for a general positive answer to my questions as "tameness" seems to be an essential part of the argument of Fulton-Olsson. But so far I have not been able to construct a counterexample. 

  [1]: http://math.berkeley.edu/~molsson/Picard1.pdf