= Solution
If $y=g\cdot x$, the component maps $g_i$ are invertible and satisfy $g_{t(\rho)}f_\rho=f'_\rho g_{s(\rho)}$. Thus they define a <quiver representation isomorphism> $X_x\to X_y$. Conversely, any such isomorphism consists of invertible maps $g_i$, and its commuting squares rearrange to $f'_\rho=g_{t(\rho)}f_\rho g_{s(\rho)}^{-1}$. Therefore
$$
\boxed{\mathcal O_X=\{x\in\operatorname{Rep}_Q(\mathbf n):X_x\cong X\}.}
$$
The <base change action on quiver representations> consequently identifies its orbits exactly with isomorphism classes at fixed <dimension vector of a quiver representation>.
Back to article page
