I somehow just realized that although I've now know the answer for a while, I had failed to give the answer to this question...

In fact, for (any) field of characteristic $p>0$, there are (plenty of) examples of reductive groups $G$ which act by group automorphisms on a vector group $U$ for which the action is not linear -- i.e. there is no $G$-equivariant isomorphism $\operatorname{Lie}(U) \simeq U$.

For an example, see < my preprint @ gmcninch.math.tufts.edu> (especially section 5). Given a non-split extension $$(\flat) \quad 0 \to W \to E \to V \to 0$$ of finite dimensional $G$-modules, one can view the extension as being defined by a suitable sort of cocycle. One can then "twist" this cocycle by Frobenius, yielding a vector group $\tilde E$ on which $G$ acts by group automorphisms.

As a $G$-module, the Lie algebra $\operatorname{Lie}(\tilde E)$ is isomorphic to $V \oplus W^{(1)}$, where the exponent on $W$ denotes the first Frobenius twist.

By construction, there is a $G$-equivariant surjection $\pi:\tilde E \to V$, and the kernel of $\pi$ identifies with $W^{(1)}$. But there is no $G$-equivariant homomorphism of algebraic groups $V \to \tilde E$ which is a section to $\pi$, hence there can be no $G$-equivariant isomorphism between $\tilde E$ and $\operatorname{Lie}(\tilde E)$.

Since there are plenty of non-split extensions $(\flat)$, there are plenty of examples...

On a more positive note, let $V$ be a vector group on which $G$ acts. The main result of the preprint just mentioned gives a sufficient condition for the linearity of an action of $G$ on $V$, and it implies that $V$ has always a filtration where the factors are vector groups with linear $G$-action, at least under the assumption that the unipotent radical of $G$ is defined and split over the ground field.

I should mention that roughly at the same time as my work on this matter, Dave Stewart found a proof of the existence of such a filtration of $V$ -- see < his paper >. His argument is simpler than mine, but it does not give a criteria for linearity, and it does not seem to work when $k$ is imperfect. (Well, his paper only states the result for $k$ algebraically closed, but I'm reasonably certain his argument is fine for perfect $k$).

I somehow just realized that although I've now know the answer for a while, I had failed to give the answer to this question...

In fact, for (any) field of characteristic $p>0$, there are (plenty of) examples of reductive groups $G$ which act by group automorphisms on a vector group $U$ for which the action is not linear -- i.e. there is no $G$-equivariant isomorphism $\operatorname{Lie}(U) \simeq U$.

For an example, see < my preprint @ gmcninch.math.tufts.edu > (especially section 5). Given a non-split extension $$(\flat) \quad 0 \to W \to E \to V \to 0$$ of finite dimensional $G$-modules, one can view the extension as being defined by a suitable sort of cocycle. One can then "twist" this cocycle by Frobenius, yielding a vector group $\tilde E$ on which $G$ acts by group automorphisms.

As a $G$-module, the Lie algebra $\operatorname{Lie}(\tilde E)$ is isomorphic to $V \oplus W^{(1)}$, where the exponent on $W$ denotes the first Frobenius twist.

By construction, there is a $G$-equivariant surjection $\pi:\tilde E \to V$, and the kernel of $\pi$ identifies with $W^{(1)}$. But there is no $G$-equivariant homomorphism of algebraic groups $V \to \tilde E$ which is a section to $\pi$, hence there can be no $G$-equivariant isomorphism between $\tilde E$ and $\operatorname{Lie}(\tilde E)$.

Since there are plenty of non-split extensions $(\flat)$, there are plenty of examples...

On a more positive note, let $V$ be a vector group on which $G$ acts. The main result of the preprint just mentioned shows that if $G$ is connected, if the unipotent radical of $G$ is defined over the ground field, and if $\operatorname{Lie}(V)$ is a simple $G$-module, then the action of $G$ on $V$ is linear. In particular, it follows (for $G$ as above) that if $G$ acts on a split unipotent group $U$, there is a filtration of $U$ by closed $G$-stable subgroups such that the quotients are vector groups on which $G$ acts linearly.