Are schematic fixed-points of a Cohen-Macaulay scheme Cohen-Macaulay? I'm not sure how long this iterative questions can go on, but let me try again.  Let's say $X$ is a Cohen-Macaulay scheme with an action of $\mathbb{G}_m$ (i.e. if $X$ is affine, a grading on the coordinate ring).   Are the schematic fixed points $X^{\mathbb{G}_m}$ of $X$ Cohen-Macaulay?  
 A: Here is a counterexample. Consider the action of $\mathbb G_{\rm m}$ on $\mathbb A^4$ defined by $t \cdot(x,y,z,w) = (x, y, tz, t^{-1}w)$, and let $X$ be the invariant closed subscheme with ideal $(xy, y^2 + zw)$; this is a complete intersection, hence it is Cohen-Macaulay. The fixed point subscheme is obtained by intersecting with the fixed point subscheme in $\mathbb A^4$, which is given by $z = w = 0$; hence it is the subscheme of $\mathbb A^2$ given by $xy = y^2 = 0$, which is of course the canonical example of a non Cohen-Macaulay scheme.
Developing this idea a little, one can show that any kind of horrible singularity can appear in the fixed point subscheme of a $\mathbb G_{\rm m}$-action on a complete intersection variety.
A: Edit: the following does not answer Ben's question. It gives an example of the subring fixed by $G_m$ being not CM, while the question asked about the subscheme of fixed points, see the comments for more details.  
Let $R$ be the (homogenous) cone of a  curve $C$ of genus $g>0$, for example $R=\mathbb C[x,y,z]/(x^3+y^3+z^3)$. Let $S=R[u,v]$, $X=\text{Spec}(S)$ and $G_m$ acts by 
$a.(x,y,z,u,v) = (ax,ay,az,a^{-1}u, a^{-1}v)$. 
Then $A= S^{G_m}$ would be a homogenous coordinate ring for $Y= C\times \mathbb P^1$, so it is not Cohen-Macaulay (if $A$ is CM, it would mean that $H^1(Y,\mathcal O_Y)=0$, impossible, see here for an explanation).
(I learned this idea from Hochster, let me try to find a reference)
