Thanks to those who clarified the question. The answer by quasi-coherent is correct. Here is one explicit example: let $G$ be $\mathbf{G}_m$, and let $V$ be a second copy of $\mathbf{G}_m$. For an integer $n>1$ (prime to the characteristic), define the action $\sigma:G\times V \to V$ by $\sigma(s,t) = s^nt$. Define $x$ to be the element $1$ in $\mathbf{G}_m$. The stabilizer (of every orbit) is $\mathbf{\mu}_n$. The quotient by the stabilizer is $q:G\to V$, $q(s) = s^n$. This is a finite, étale morphism. The fiber product $G\times_V G$ as a closed subscheme of $G\times G$ is the image of the closed immersion $$i:\mathbf{\mu}_m\times \mathbf{G}_m\to \mathbf{G}_m \times \mathbf{G}_m, \ (r,s) \mapsto (rs,s).$$ The identity $\text{Id}_G:G\to G$ gives an element $[\text{Id}_G]\in G(x)(G)$. Also the two pullbacks to $G\times_V G$ are equal. So, were your functor representable and thus a sheaf for the étale topology, then $[\text{Id}_G]$ would be the pullback of some $[g]\in G(x)(V)$. In turn, this would be an equivalence class of morphisms $g:\mathbf{G}_m\to \mathbf{G}_m$ such that $q\circ g$ equals the identity. However, there is no such morphism $g$. Therefore your functor is not an étale sheaf. Therefore it is not representable.

twobackslashes first. – MTS Aug 16 '12 at 14:08