 Solution

If $y=g\cdot x$, the component maps $g_i$ are invertible and satisfy $g_{t(\rho )}{f}_{\rho }=f_{\rho }^{\prime }{g}_{s(\rho )}$. 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 $f_{\rho }^{\prime }=g_{t(\rho )}{f}_{\rho }{g}_{s(\rho )}^{-1}$. ThereforeThe base change action on quiver representations consequently identifies its orbits exactly with isomorphism classes at fixed dimension vector of a quiver representation.