 Solution

A unipotent algebraic group admits a faithful linear representation in which every group element is a unipotent matrix. For $E={\mathrm{End}}_Q(X)$, invertibility is the nonvanishing condition ${\prod }_i\det h_i\ne 0$. Therefore ${\mathrm{Aut}}_Q(X)=E^{\times }$ is a nonempty Zariski-open subset of the vector space $E$.

Take a Krull-Schmidt decomposition $X\cong {⨁}_{a=1}^r{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposable modules. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local endomorphism ring. Its residue division algebra is $k$: over an algebraically closed field, every element of a finite-dimensional division algebra has an eigenvalue and hence must be scalar. The semisimple quotient of a module endomorphism algebra consequently gives, for the Jacobson radical $J=J(E)$,is surjective with kernel $1+J$. The nilpotence of $J$ makes $(1+j{)}^{-1}=1-j+j^2-\cdots$ finite and each $1+j$ unipotent. The kernel is closed and normal. Acting on the multiplicity spaces embeds the product of general linear groups back into $E^{\times }$ and splits this quotient. This proves the Levi decomposition of a quiver automorphism groupSince $1+J$ is the unipotent radical, a nonzero $X$ is indecomposable exactly when ${\mathrm{Aut}}_Q(X)/R_u({\mathrm{Aut}}_Q(X))\cong {\mathbb{G}}_m$: the product has a single factor of size one.

For the base change action on quiver representations, the orbit map is $g\mapsto (g_j{x}_{\rho }{g}_i^{-1}{)}_{\rho :i\to j}$. Substituting $g_i=I+ϵu_i$, with ${ϵ}^2=0$, shows its differential isIts kernel is ${\mathrm{End}}_Q(X)$. The stabilizer is smooth because it is open in that vector space. Hence the differential has rank $\dim \mathrm{GL}(\mathbf{n})-\dim {\mathrm{Aut}}_Q(X)=\dim {\mathcal{O}}_X$, and its image is the Zariski tangent space $T_x{\mathcal{O}}_X$. The normal space to a quiver orbit is thereforeThe ambient quiver representation space is an irreducible affine space, and orbits are locally closed. An orbit is open exactly when its dimension equals that ambient dimension, equivalently when ${\mathrm{Ext}}_Q^1(X,X)=0$. This proves that rigid quiver representations have open orbits. Such an orbit is dense and unique, since two nonempty open subsets of an irreducible space intersect.