Let $X$ be a quasi-projective scheme over a field $k$.
Let $G$ be a finite group acting on $X$ whose order is invertible in $k$.
If $X$ is Cohen-Macaulay, can we conclude that the subscheme of fixed points $X^G$ is Cohen-Macaulay?



Fixed point scheme of finite group Cohen-Macaulay?
Here is a simpler example than the one I left before, using the same strategy. Let
$$X = \{ x\_1 x\_3 = x\_1 x\_4 = x\_1 x\_5 = x\_2 x\_4 = x\_2 x\_5 = x\_3 x\_5 = 0 \} \subset \mathbb{C}^5.$$
This is the reduced union of four $2$-planes. Here is a projective picture, where $j$ represents the point where $x\_j$ is the sole nonzero coordinate:
$$1 - 2 - 3 - 4 - 5.$$
The graph above is shellable, so this is Cohen-Macaulay.


Now, let $C\_2$ act on $X$ by $(x\_1, x\_2, x\_3, x\_4, x\_5) \mapsto (x\_1, x\_2, - x\_3, x\_4, x\_5)$. Then the fixed locus of $C\_2$ (even scheme-theoretically) is
$$Y = \{ x\_1 x\_4 = x\_1 x\_5 = x\_2 x\_4 = x\_2 x\_5 = x\_3 = 0 \}.$$


This is the reduced union of two $2$-planes; we can visualize it as
$$1 - 2 \phantom{- 3 -} 4 - 5.$$
That is a standard example of a non-Cohen-Macaulay ring.


