Let us suppose the the group $G:=\mathbb{Z}/2\mathbb{Z}=(1,i)$ freely act on a smooth projective variety/$k$ $X$ and denote by $Y$ the G.I.T. quotient $X/G$. Let $\pi:X\longrightarrow Y$ the quotinet map. Now take a $G$-linearised coherent sheaf $(\mathcal{F}, \lambda)$, one can construct the sheaf of invariants of $\mathcal{F}$,the sheaf $\pi_\ast(\mathcal{F})^G$, that  is a sheaf on $Y$. Then given $\overline{\alpha}=\pi(\alpha)$ a point of $Y$ it is possible to consider the fiber of  $\pi_*(\mathcal{F})^G$ at $\overline{\alpha}$, $\pi_\ast(\mathcal{F})^G(\overline{\alpha})$. On the other and one can consider the fiber of $\pi_\ast(\mathcal{F})(\overline\alpha)$ that is (I think) isomorphic to the direct sum
$$\mathcal{F}(\alpha)\oplus\mathcal F(i\cdot\alpha)\simeq \mathcal F\otimes(k(\alpha)\oplus k(i\cdot\alpha)$$
Now this vector space admit a $G$-action induced by $\lambda$, given by $\lambda\otimes\sigma$ where sigma is the action on $k(\alpha)\oplus k(i\cdot\alpha)$ given by permutation.

My question is: the fiber of the sheaf of invariants is isomorphic to the invariant subspace of the fiber (in the given action)?
I think the answer is no (the first is a subspace of the latter), but I would prefer it to be yes...

If the answer is yes, how do I prove it? could you give me some references?

If the answer is no, could you give me an explicit counterexample?

Thank you very much for the time you dedicated to me and a special thanks to every one who will answer me
best regards
Stgermain