I don't know the answer, but here's a question that might lead someone to an answer. Let's suppose that we're working over an algebraically closed field $k$ of characteristic $p$. Let $Fr$ denote the Frobenius automorphism.
From a recent paper of Tanaka and Kaneta (Hiroshima Math. J. 2008), I found that $Aut(G_a^n)$ (the automorphism group of the vector group $G_a^n$ over $k$) can be identified with: $$\{ A \in M_n(k)[[F]]^\times : A,A^{-1} \in M_n(k)[F] \}.$$ Here, we work in a ring of noncommutative power series in the formal variable $F$, with coefficients in the matrix ring $M_n(k)$, subject to the natural relation: $$F \cdot m = Fr(m) \cdot F.$$ I.e., one may pass all F's to the right, by applying Frobenius to the entries of the matrices.
Now, there is a natural surjective homomorphism $lin$ from $Aut(G_a^n)$ to $GL_n(k)$, obtained by taking the "constant term" of the power series.
What does the kernel of $lin$ look like? Is it (representable by?) a pro-unipotent group scheme over $k$? I have no idea.
Though it's a bit obvious, the "Refined Question" would be answered if one could prove the following: If a group $G$ acts on the vector group $G_a^n$ via $\alpha: G \rightarrow Aut(G_a^n)$, then $Ker(lin \circ \alpha)$ is a unipotent subgroup of $G$.