Let $k$ be a field; I'm going to discuss linear algebraic groups over $k$. The question I'll pose is only interesting when the characteristic is $p>0$.

*1. Some motivation*

A *vector group* is an algebraic group isomorphic (over $k$) to a product of (finitely many) copies of the additive group $\mathbf{G}_a$. Let $U$ be any connected unipotent group over $k$. If $k$ is perfect, or if $U$ is $k$-split, then $U$ has a filtration
$1 = U_0 \subset U_1 \subset \cdots \subset U_n = U$
where each $U_i$ is normal in $U$ and each $U_i/U_{i-1}$ is a vector group. If $U$ is a normal subgroup of a linear group $G$, you can
arrange that each $U_i$ is invariant under conjugation by $G$. This suggests that to study the *group extension* $$(*) \quad 1 \to U \to G \to G/U \to 1,$$ one might profitably study first the case where $U$ is a vector group.

Let $G$ be a linear group and suppose that the unipotent radical $R$
of $G$ is defined over $k$ and is $k$-split (each of these conditions
can fail in general; they always hold when $k$ is perfect). Then the
question of whether $(*)$ splits when $U=R$ is precisely the question
of whether $G$ has a *Levi factor*; cf. this question of Jim Humphreys.

*2. Action on a vector group*

Let $U$ be a vector group and suppose that the linear group $G$ acts on $U$ by algebraic group automorphisms. The action of $G$ on $U$ determines an action of $G$ on $\mathfrak{u}=\operatorname{Lie}(U)$.

Question (first approximation):Is there a $G$-equivariant isomorphism $\mathfrak{u} \to U$ (where the vector space $\mathfrak{u}$ is viewed as a vector group in the obvious fashion)?

I'll say that the action of $G$ on $U$ is *linearizable* if
there is such an equivariant isomorphism.

Some remarks: If the action of $G$ on $U$ is linearizable, then $G$ centralizes the action of the multiplicative group $\mathbf{G}_m$ on $U$ obtained by transport of structure from scalar multiplication on $\mathfrak{u}$. This $\mathbf{G}_m$-action determines a grading on the algebra $k[U]$ of regular functions on $U$ which is stable for the action of $G$ on $k[U]$.

*3. Partial answers*

*Postive:* If the characteristic of
$k$ is $0$, the above question has always an affirmative answer. (Use the exponential mapping $\mathfrak{u} \to U$; this is $G$ equivariant e.g. because there is a faithful linear representation of the semidirect product $G \ltimes U$).

*Negative:* If $G = \mathbf{G}_a$, the
question has a negative answer in char. $p>0$. Consider the action of
$G$ on $V = {\mathbf{G}_a}^2$ given by the rule $$x.(a,b) = (a +
xb^p,b).$$
The action of $G$ on $\operatorname{Lie}(V)$ is trivial, so there is no
$G$-equivariant isomorphism $\operatorname{Lie}(V) \to V$.

*4. What I meant to ask.*

Refined question:If $G$ is reductive, is the action of $G$ on a vector group $U$ always linearizable?

An affirmative answer would mean that when $G/U$ is reductive, one can study the group extension $(*)$ using cohomology of linear representations of $G/U$.

Note that are known examples even in char. 0 (Schwarz, Kraft,...) of actions of reductive $G$ on affine space $A=\mathbf{A}^n$ which are not linear, but, at least in char. 0, these actions don't respect a vector group structure on $A$.