TITLE: Whitney stratification of algebraic varieties
QUESTION [5 upvotes]: When do the orbits of an action on an algebraic variety make a Whitney stratification?

REPLY [2 votes]: The answer is indeed always when there are fnitely many orbits. Let $Y \subset X$ be an orbit, $y \in Y$, and $U \subset X$ a $G$-invariant open such that $y \in Y \cap U$, the relative open $Y \cap U$ is dense in $Y$, closed in $U$ and $U$ embedds $G$-equivariantly in some $\mathbb{C}^N$.
Let us denote by:
$$ C(X) = \overline{\{(x,H) \in U_{smooth} \times (\mathbb{C}^N)^*, \ \textrm{such that} \ T_{X,x} \subset H \}},$$
where the overline is the Zariski closure in $\mathbb{C}^N \times (\mathbb{C}^N)^*$. The variety $C(X)$ is known as the conormal space and is also $G$-invariant. Note that the projection map : $\pi : C(X) \longrightarrow X$ is $G$-equivariant. Let:
$$ \rho : \mathrm{Bl}_{\pi^{-1}(Y)}C(X) \longrightarrow C(X)$$
be the blow-up of $C(X)$ along $\pi^{-1}(Y)$. The map $\rho$ is again $G$-equivariant, so that the composite:
$$ \pi \circ \rho : \mathrm{Bl}_{\pi^{-1}(Y)}C(X) \longrightarrow C(X)$$
is $G$-equivariant. In particular, all fibers of  $\pi \circ \rho$ are isomorphic and we find :
$$ \dim (\pi \circ \rho)^{-1}(y) = \dim (\pi \circ \rho)^{-1}(Y) - \dim Y = N-2 - \dim Y.$$
Hence, proposition 2.3.1 in the paper Variétés Polaires II by Bernard Teissier shows that the pair $(X,Y)$ staisfies the Whitney conditions at $y$. This is true for all $y \in Y$, so that the pair $(X,Y)$ satisfies the Whitney conditions.
As hinted in the comments, if there are finitely many orbits, then you get a finite stratification of $X$ by orbits, and by the above arguments, this is a Whitney stratification.
Note however that it is certainly not the minimal stratification, as illustrated by the case $X$ smooth and $G$ acts non-trivially with finitely many orbits.