TITLE: Do orbits and stable loci of group actions have natural scheme structures?
QUESTION [14 upvotes]: Suppose G is an algebraic group with an action G×X→X on a scheme. Then many of the usual constructions you make when you talk about group actions on sets can be made scheme-theoretically. For example, if x∈X is a point (thought of as a map x:∗→X, where ∗ is Spec of a field or the base scheme), then the stabilizer Stab(x) is naturally a scheme because it is the fiber product

Stab(x) ----> G×X    (g,y)
  |            |       _
  |            |       |
  v   (x,x)    v       v
  ∗ --------> X×X   (gy,y)


Does the orbit of a point have a natural scheme structure?
Does the fixed locus (the set of points x∈X fixed by all of G) have a scheme structure?

For (1), if everything is sufficiently nice, then the morphism G×∗→X, given by g→g⋅x has a scheme-theortic closed image, and the actual image is constructible and invariant under the G-action, so the actual image is an open subset of its closure. Thus, the orbit gets the structure of an open subscheme of a closed subscheme of X. But this construction doesn't feel very natural.
For (2), you can obviously define the functor Fix(T)={t∈X(T)|t is fixed by every element of G(T)}. Is this functor always representable?
Edit: Given that Scott has given such an excellent (negative) answer to question (1) but not said anything about question (2), I've asked (2) as a separate question.

REPLY [8 votes]: Question number 1: Let your base S be Spec k[x] (say k is an algebraically closed field), let X be Spec k[x,y], and let G be Ga,S, with action over the point s given by gs(xs) = (sgs) + xs.  This action is transitive away from zero, so the orbit of the zero section is a plane with a slit.  This is not an open subscheme of a closed subscheme of X, because it is not a scheme.