Is it possible to have a point $P_1$ not $\chi$-semistable but $P_2$ $\chi$-semistable with these two points in the same orbit?

Let $G$ be a group acting on an affine variety $X\subseteq \mathbb{A}_{\mathbb{C}}^n$.

Suppose $P_1$ and $P_2$ are two points in $X$ such that $g\circ P_1=P_2$ for some $g\in G$. This means that $P_2$ is in the $G$-orbit of $P_1$, and vice versa.

Let $\chi$ be a character of $G$.

We say $P$ is $\chi$-semistable if there exists a function $f:X\rightarrow \mathbb{C}$ so that $f(g\circ P)=\chi(g)f(P)$ for all $g\in G$ with $f(P)$ nonvanishing.

$\mathbf{Question}:$ Is it possible for $P_1$ to not be $\chi$-semistable but $P_2$ is $\chi$-semistable? That is, is it possible to have the following: there doesn't exist a $\chi$-semistable function for $P_1$ but there exists a $\chi$-semistable function for $P_2$ even though they are in the same $G$-orbit?

I am currently having a somewhat difficult time coming up with such an example (I'm trying to come up with such example by taking $G=\mathbb{C}^{\times}$ or $GL_n(\mathbb{C})$ acting on some affine $n$-space).

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No, any $\chi$-semistable function $f$ for $P_1$ is also one for $P_2$, and vice versa. Just compute and verify this, using that $\chi$ is a character.
Thank you Dan! So the same $\chi$-semiinvariant function for $P_1$ works for all points in the orbit $G\circ P_1$? Must the orbit be closed, or this doesn't matter? And does the $\chi$-semiinvariant function extend to orbit closures? – math-visitor Jul 17 '12 at 8:50