Let $G$ be a linearly reductive algebraic group, and let $X$ be an irreducible affine variety, over an algebraically closed field $\mathbb{K}$, with a regular action of $G$. By Rosenlicht's result, we know that there's a $G$-invariant open set $U\subseteq X$ such that the geometric quotient $\Phi:U\rightarrow U/\!/G$ exists. So, "generic" orbits are contained in $U$. But is there something more specific about these "generic" orbits known?
My question:
This $U$ contains orbits of maximal dimension, but is there more specific information for which $x\in X$, the orbit $\mathcal{O}(x)\subset U$?
I'm interested in the setting of quiver representations:
Let $\mathbb{K}$ be an algebraically closed field of characteristic $0$. Let $A$ be a finite-dimensional (associative and unital) algebra over $\mathbb{K}$. Assume there is a quiver $Q=(Q_0,Q_1)$, where $Q_0$ is the set of vertices and $Q_1$ is the set of arrows, associated to $A$, i.e., $A\cong\mathbb{K}Q/I$, with $0 \neq I\subset \mathbb{K}Q$ an admissible ideal. Further assume that the algebra $A$ is such that the quiver $Q$ is acyclic.
For a dimension vector $\beta$, the representation space $$\DeclareMathOperator\rep{rep}\DeclareMathOperator\GL{GL} \rep_{\beta}(Q,I):=\biggl\lbrace M\in\prod_{a\in Q_1}\operatorname{Mat}_{\beta(ha)\times\beta(ta)}(\mathbb{K}) \;\bigg|\; \text{$M(r) = 0$ for all $r \in I$}\biggr\rbrace $$ parametrizes $\beta$-dimensional representations of $(Q,I)$. The linear algebraic group $$ \GL_{\beta}:=\prod_{x\in Q_0}\GL_{\beta(x)}(\mathbb{K}) $$ acts on $\rep_{\beta}(Q,I)$ by change of basis. Now take an irreducible component $\mathcal{C}\subseteq\rep_{\beta}(Q,I)$ and restrict the action of $\GL_{\beta}$ to $\mathcal{C}$.
So my question now becomes:
Is there any information for which representations $V\in\mathcal{C}$, the orbit $\mathcal{O}(V)\subset U$, other than saying $V$ has to be in "general" position?