= Solution
An <algebraic group> is <connected> when its underlying <Zariski topology> is a <connected space>. An <unipotent algebraic group> is a <linear algebraic group> whose elements, in a faithful matrix realization, satisfy that $u-I$ is nilpotent. A <reductive algebraic group>, in the connected convention, is a <connected> <linear algebraic group> whose <unipotent radical> is trivial: it has no nontrivial connected normal unipotent algebraic subgroup.
Let $E=\operatorname{End}_Q(X)$. It is a finite-dimensional associative algebra, and $\operatorname{Aut}_Q(X)=E^\times$. The latter is the nonempty open subset where the determinant on $\bigoplus_iX_i$ is nonzero. Since the affine space $E$ is irreducible, this open subset is irreducible and hence connected. It is a <linear algebraic group>: within $\operatorname{GL}(\bigoplus_iX_i)$ it is cut out by the linear equations of preserving the vertex summands and commuting with the arrow maps. Thus \b[the automorphism group is connected].
By the <Krull–Schmidt theorem>, write $X\cong\bigoplus_{a=1}^sM_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposables $M_a$. The <Fitting lemma> makes each $\operatorname{End}_Q(M_a)$ a local algebra. Its residue division algebra is $k$: every element of a finite-dimensional division algebra over the algebraically closed field has an eigenvalue for left multiplication, and subtracting that scalar gives a noninvertible element, hence zero.
The standard radical description of the endomorphism algebra of a <Krull-Schmidt decomposition> therefore gives
$$
E/J(E)\cong\prod_{a=1}^s\operatorname{Mat}_{m_a}(k).
$$
Concretely, maps between nonisomorphic summands lie in the radical, while on each isotypic block one reduces all entries modulo the local radical. The off-diagonal rule follows because a composition $M_a\to M_b\to M_a$ cannot be a unit for $a\ne b$: it would split $M_a$ off the indecomposable $M_b$, forcing an isomorphism. The finite-dimensional radical description then identifies the kernel of this block reduction with $J(E)$. Put $U=1+J(E)$. The <Jacobson radical> of the finite-dimensional algebra is nilpotent, so every $1+j$ is a unit and is unipotent on $\bigoplus_iX_i$. Also $U$ is closed, being the translate of the linear subspace $J(E)$, and normal, being the kernel of
$$
E^\times\longrightarrow\prod_a\operatorname{GL}_{m_a}(k).
$$
This map has an explicit algebraic section: after fixing the direct-sum decomposition, a matrix $B_a$ acts on the $m_a$ copies by $B_a\otimes1_{M_a}$. These block actions give a subgroup $L\cong\prod_a\operatorname{GL}_{m_a}(k)$ with trivial intersection with $U$. Every unit is uniquely a product of an element of $U$ and one of $L$. Hence
$$
\boxed{\operatorname{Aut}_Q(X)\cong(1+J(E))\rtimes\prod_{a=1}^s\operatorname{GL}_{m_a}(k).}
$$
This <Levi decomposition of a quiver automorphism group> holds in every characteristic; the section is constructed directly from the multiplicity spaces.
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