connected components of the fixed point subvariety Let $X$ be a smooth complex variety with an action of a finite group $G$.
The fixed point subvariety $X^G$ is smooth but may have many connected components.
What determines these connected components geometrically? For example it seems possible to argue that the induced action of $G$ on the tangent spaces of various points on any connected component are all isomorphic as $G$ representations. Does this determine the connected components? More precisely suppose $x,y \in X^G$ and suppose $T_x X \simeq T_yX$ as $G$-representations, then do they lie in the same connected component of $X^G$?
 A: No. Consider $X=\mathbb{CP}^1 \times \mathbb{CP}^1$ with the $\mathbb{Z_{k}}$-action generated by the map $$\mu. ([x_0:x_1],[y_0:y_1]) =  ([\mu x_0:x_1],[\mu y_0:y_1]),$$ for $\mu = e^{\frac{2\pi i}{k}}$.
Then there are exactly four fixed points $p_0=([1:0],[1:0])$,$p_1=([0:1],[1:0])$,$p_2=([1:0],[0:1])$,$p_3=([0:1],[0:1])$.
The $\mathbb{Z}_{k}$-representations on $T_{p_{1}}(X)$ and $T_{p_{2}}(X)$ are isomorphic. Both are the representation on $V=\mathbb{C}^2$ $\mu.(z_1,z_{2}) = (\mu z_1, \mu^{-1} z_{2})$ in suitable complex coordinates (just choose suitable affine co-ordinates on the $\mathbb{CP}^1$ factors).
Edit:
For the general question, there is one cautionary example that one should be aware of. Consider the $\mathbb{Z}_{k}$-actionon $\mathbb{CP}^n$, $$\mu. [z_0: \ldots :z_{n}] = [\mu z_0: \ldots : z_n].$$ The fixed point set consists of the hyperplane $H = \{z_0=0\} \cong \mathbb{CP}^{n-1}$ and $p= [1:0: \ldots :0]$.
Next, if $Y \subset H$ is any smooth projective subvariety. Then, one can blow-up $Y$ $\mathbb{Z}_{k}$-equivariantly, since it contained in the fixed point set of the original action. Then, we have a $\mathbb{Z}_{k}$-action on $Bl_{Y}(\mathbb{CP}^n)$ for which the fixed set consists of, a point, a copy of $\mathbb{CP}^{n-1}$ and a copy of $Y$.
This is quite cautionary, becuase $Bl_{Y}(\mathbb{CP}^n)$ is has several geometrically rare properties, for example it is rational. But, $Y$ can be taken to be any smooth projecitve variety, hence there isn't really any geometric properties that carry over to all of the fixed components. Although, in the case that the group action extends to a algebraic group action, there are results that say that some of the fixed components inherit geometric properties of $X$, see papers of Bialynicki-Birula and others.
