Let $X/k$ be a scheme of finite type over a field $k$, $G/k$ be a finite group scheme and suppose $G$ acts on $X$, i.e we have a $k$-morphism $ \mu : G \times_k X \rightarrow X$ satisfies some conditions, see Mumford's book "Abelian Varieties", p. 108.

Suppose that $k$ is algebraically closed and consider the map $\mu$ induced on closed points $G(k) \times X(k) \rightarrow X(k)$, which gives $G(k) \rightarrow \mathrm{Aut}(X(k))$. Under the assumption that each $x \in X$ has an open affine neighborhood $U$ such that $U$ is invariant under $G$, one can form the quotient $X/G$. In particular, quasi-projective varieties have this property. This assumption reduces the construction from general variety to affine case.

For a general field $k$, if one can cover $X$ by open affine subset $U_i$ such that the image of $ G \times_k U_i$ under $\mu$ is contained in $U$, then one reduces the construction to the affine case. But I couldn't figure out if a quasi-projective variety $X$ over $k$ always has such a covering. By working with base change to the algebraic closure of $k$, one gets a $G$-invariant open affine covering of $X_{\overline{k}}$. The images of these open affine subsets of $X_{\overline{k}}$ under the projection to $X$ is still $G$-invariant, but not necessary affine. So for quasi-projective varieties over $k$, do we have the existence of the quotient $X/G$?

Another question is that if it exists, then if it is also quasi-projective? What I know is that, in the classical case ( over algebracially closed field and giving finite group action $H \rightarrow \mathrm{Aut}_k (X))$, if the quotient exists and $X$ is complete, then $X/H$ is also complete.