TITLE: Algebraic proof without using comparison theorem for étale cohomology
QUESTION [7 upvotes]: Let $X$ be some smooth scheme over $\mathbf C$ equipped with an action of $\mu_n$ (the group of $n$th roots of unity). 
The étale cohomology groups of X are therefore equipped with an action of $\mu_n$.
Now, let's suppose that the action of $\mu_n$ on $X$ extends to an action of $\mathbf G_m$.
Then, the analytic space $X(\mathbf C)$ is equipped with an action of $\mu_n$ which extends to an action og $\mathbf C^\times$. As $\mathbf C^\times$ is connected, one concludes that the action of $\mu_n$ on the singular cohomology of $X(\mathbf C)$ is trivial (the multiplication by each element $\xi\in \mu_n$ is homotopic to the identity).
Going backwards to étale cohomology, using a comparison theorem between étale and singular cohomology, one concludes that the action of $\mu_n$ on the étale cohomology of $X$ is trivial.
Is there a direct proof  that the action on the étale cohomology groups is trivial without having to refer to the topological case?

REPLY [4 votes]: Here's an alternative to grghxy's answer. Let me stick to the original assumptions for simplicity, but they can certainly be relaxed. Let $G$ be the image of the $\mu_n$ action on étale cohomology with finite coefficients. In particular, $G$ is finite. Let $g\in G$. Since the action is assumed to extend to a $\mathbb{C}^*$-action, we can solve $g=h^N$ for arbitrary $N>0$. Now choose $N=|G|$. 
NB  See some comments by grghxy & Yonatan Harpaz below.