Is the orbit of a finite set of algebraic numbers under the action of the absolute galois group finite?

Fix an algebraic closure $\bar{\mathbf{Q}}$ of $\mathbf{Q}$.

Let $B\subset \mathbf{P}^1_{\bar{\mathbf{Q}}}$ be a closed subscheme of finite cardinality.

Let $K$ be a number field such that $B$ can be defined over $K$. Let $B_K$ be a closed subscheme of $\mathbf{P}^1_K$ such that the base change to $\bar{\mathbf{Q}}$ is $B$.

Is the orbit of $B_K$ under the action of the absolute Galois group Gal$(\bar{\mathbf{Q}}/\mathbf{Q})$ finite? Is it a closed subscheme?

The answer is trivially yes if $K=\mathbf{Q}$. I expect the cardinality of the orbit to be less or equal to $[K:\mathbf{Q}]\cdot$#$B$ in general. But why?

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You expect something and are asking us why? (last line of the question) maybe you could tell us that better. =P – Patrick Da Silva Jul 29 '11 at 20:46
The title asks about the orbit of an algebraic number, but the body asks about the orbit of a closed subscheme (of ${\bf P}^1_K$, etc.). If these are not the same, perhaps some editing is in order to bring title and body into alignment. – Gerry Myerson Jul 29 '11 at 23:11
@Patrick. By "I expect that " I mean that "I would not be surprised if"...In any case, I might be wrong. I just thought it would be nice to add some personal intuition. – Oen Jul 30 '11 at 0:33
@Gerry. There's not a big difference in the two questions. I just tried to minimize the length of the title. – Oen Jul 30 '11 at 0:34
@Oen: I don't know any arithmetic geometry, but could you explain why you are considering the projective line $\mathbf{P}^1_{\overline{\mathbf{Q}}}$ instead of just, say, $\operatorname{Spec} \overline{\mathbf{Q}}[x]$? Does the presence of points at infinity simplify some issue under consideration? – Zhen Lin Jul 30 '11 at 9:21