= Solution
A <unipotent algebraic group> admits a faithful linear representation in which every group element is a <unipotent matrix>. For $E=\operatorname{End}_Q(X)$, invertibility is the nonvanishing condition $\prod_i\det h_i\ne0$. Therefore $\operatorname{Aut}_Q(X)=E^\times$ is a nonempty <Zariski-open subset> of the vector space $E$.
Take a <Krull-Schmidt decomposition> $X\cong\bigoplus_{a=1}^rM_a^{\oplus m_a}$, with pairwise nonisomorphic <indecomposable modules>. The <Fitting lemma> makes each $\operatorname{End}_Q(M_a)$ 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> $J=J(E)$,
$$
E/J\cong\prod_{a=1}^rM_{m_a}(k),\qquad E^\times\longrightarrow\prod_{a=1}^r\operatorname{GL}_{m_a}(k)
$$
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 group>
$$
\boxed{\operatorname{Aut}_Q(X)\cong(1+J)\rtimes\prod_{a=1}^r\operatorname{GL}_{m_a}(k)}.
$$
Since $1+J$ is the <unipotent radical>, a nonzero $X$ is indecomposable exactly when \b[$\operatorname{Aut}_Q(X)/R_u(\operatorname{Aut}_Q(X))\cong\mathbb G_m$]: the product has a single factor of size one.
For the <base change action on quiver representations>, the orbit map is $g\mapsto(g_jx_\rho g_i^{-1})_{\rho:i\to j}$. Substituting $g_i=I+\epsilon u_i$, with $\epsilon^2=0$, shows its differential is
$$
\boxed{\xi_x(u)_\rho=u_jx_\rho-x_\rho u_i}.
$$
Its kernel is $\operatorname{End}_Q(X)$. The stabilizer is smooth because it is open in that vector space. Hence the differential has rank $\dim\operatorname{GL}(\mathbf n)-\dim\operatorname{Aut}_Q(X)=\dim\mathcal O_X$, and its image is the <Zariski tangent space> $T_x\mathcal O_X$. The <normal space to a quiver orbit> is therefore
$$
\boxed{T_x\operatorname{Rep}_Q(\mathbf n)/T_x\mathcal O_X\cong\operatorname{Ext}^1_Q(X,X)}.
$$
The 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 $\operatorname{Ext}^1_Q(X,X)=0$. 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.
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