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[Update below: I think this answers the question now.] 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 $G$.

Update: Following David's comment, it appears that $Ker(lin)$ is a pro-unipotent group over $k$.

I think that this does it, though I'm shaky on pro-affine groups. But the image of a reductive group is reductive, even in the pro-affine setting (Section 3 of Mostow-Hochschild, Pro-Affine Algebraic Groups, Amer. J. of Math 1969. Happily this result is right before the assumption of char=0). Any reductive subgroup of a unipotent group is trivial, in this same reference. Hence the image of a reductive group in $Ker(lin)$ must be trivial.

I think this answers the question, as long as I'm not missing any subtleties in the definitions of reductive/unipotent in the pro-affine setting.

[Update below: I think this answers the question now.] 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.

Update: Following David's comment, it appears that $Ker(lin)$ is a pro-unipotent group over $k$.

The image of a reductive group is reductive, even in the pro-affine setting (Section 3 of Mostow-Hochschild, Pro-Affine Algebraic Groups, Amer. J. of Math 1969. Happily this result is right before the assumption of char=0). Any reductive subgroup of a unipotent group is trivial, in this same reference. Hence the image of a reductive group in $Ker(lin)$ must be trivial.

Update: Let $\alpha: G \rightarrow Aut(G_a^n)$ be an algebraic action of $G$ on the vector group $G_a^n$.

It follows that $H = \alpha(G)$ is a reductive subgroup of the pro-affine group $Aut(G_a^n)$, which does not intersect the pro-unipotent subgroup $U = Ker(lin)$. It remains to prove that $H$ is conjugate to the subgroup $GL_n \subset Aut(G_a^n)$.

Let $H' = lin(H)$ be the projection of $H$ in $GL_n$ -- this projection is an isomorphism from $H$ to $H'$. We find that $H \cdot U = H' \cdot U$, and $U$ is the unipotent radical of $H \cdot U$.

Now the remaining question: are Levi factors of $H \cdot U$ (geometrically, since we work over $k = \bar k$) conjugate? To this, I think I defer back to the OP, who has recently written a paper on this topic, which will hopefully adapt to this pro-affine group setting.

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 $G$.

Update: Following David's comment, it appears that $Ker(lin)$ is a pro-unipotent group over $k$.

I think that this does it, though I'm shaky on pro-affine groups. But the image of a reductive group is reductive, even in the pro-affine setting (Section 3 of Mostow-Hochschild, Pro-Affine Algebraic Groups, Amer. J. of Math 1969. Happily this result is right before the assumption of char=0). Any reductive subgroup of a unipotent group is trivial, in this same reference. Hence the image of a reductive group in $Ker(lin)$ must be trivial.

I think this answers the question, as long as I'm not missing any subtleties in the definitions of reductive/unipotent in the pro-affine setting.

[Update below: I think this answers the question now.] 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 $G$.

Update: Following David's comment, it appears that $Ker(lin)$ is a pro-unipotent group over $k$.

I think that this does it, though I'm shaky on pro-affine groups. But the image of a reductive group is reductive, even in the pro-affine setting (Section 3 of Mostow-Hochschild, Pro-Affine Algebraic Groups, Amer. J. of Math 1969. Happily this result is right before the assumption of char=0). Any reductive subgroup of a unipotent group is trivial, in this same reference. Hence the image of a reductive group in $Ker(lin)$ must be trivial.

I think this answers the question, as long as I'm not missing any subtleties in the definitions of reductive/unipotent in the pro-affine setting.

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 $G$.

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 $G$.

Update: Following David's comment, it appears that $Ker(lin)$ is a pro-unipotent group over $k$.

I think that this does it, though I'm shaky on pro-affine groups. But the image of a reductive group is reductive, even in the pro-affine setting (Section 3 of Mostow-Hochschild, Pro-Affine Algebraic Groups, Amer. J. of Math 1969. Happily this result is right before the assumption of char=0). Any reductive subgroup of a unipotent group is trivial, in this same reference. Hence the image of a reductive group in $Ker(lin)$ must be trivial.

I think this answers the question, as long as I'm not missing any subtleties in the definitions of reductive/unipotent in the pro-affine setting.