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Given a finite group $G$ acting on an algebraic variety $X$ (let's say over $\mathbb C$, if that helps), is there always a proper variety $\bar X$ with a $G$ action such that $X \to \bar X$ is a $G$-equivariant open dense embedding?

Nagata's proof of the existence of compactifications of algebraic varieties seems to depend on arbitrary choices, so the embedding $X\to \bar X$ produced in the original proof doesn't have to respect the automorphisms of $X$. In fact, one cannot hope in general to find an embedding respecting all the automorphisms of $X$, since, for example already $\Bbb A^2$ has far more automorphisms than any proper algebraic surface.

Therefore, the above statement is probably the only equivariance statement for Nagata's theorem one can hope for.

(I think, one could tackle the problem as follows, at least in the case where $G$ acts freely on $X$: Consider the quotient $X//G$, which exists as an algebraic space. According to this paper, Nagata's theorem is also true for algebraic spaces giving us a proper algebraic space $\overline{X//G}$. If one could extend the $G$-torsor on $X//G$ to $\overline{X//G}$ and show that it's total space is in fact a scheme, one would get a desired compactification.)

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    $\begingroup$ Yes, this can be done in a simple way that doesn't require any use of quotients (or freeness hypotheses on the action), by a technique which comes up (in a slightly different way) in the proof of the compactification theorem for algebraic spaces (if you read it). Let $j:X \hookrightarrow Y$ be an initial choice of compactification. Define $f:X \rightarrow Y^G := \prod_{g \in G} Y$ by $x \mapsto (j(g^{-1}x))$. Then $f$ is an immersion (exercise) and $G$-equivariant where $G$ acts on $Y^G$ via $g_0.(x_g) = (x_{g_0^{-1}g})$. Let $\overline{X}$ be the schematic image of $X$. $\endgroup$ – nfdc23 Nov 25 '15 at 13:50

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