It is well known that any unipotent algebraic group (over a field) can be embedded as a closed subgroup of $U_n$ for some $n$, where $U_n$ denotes the set of all $n \times n$ upper triangular matrices with $1$'s on the diagonal.

An algebraic subgroup $H$ of an algebraic group $G$ is called $\textbf{observable}$ if every (rational) linear representation of $H$ arises as the restriction of a (rational) linear representation of $G$. My question:

Under what conditions, if any, is a unipotent algebraic group observable with respect to its overgroup $U_n$?

Background: I have in hand a certain result concerning the characteristic $p>0$ representation theory of the groups $U_n$, when $p$ is sufficiently large when compared to $n$ and the dimension of a representation. It essentially says that such representation look functorially identical to representations of $U_n^\infty$ (countable direct product of copies of $U_n$) in characteristic zero.

If I can pass representations of a unipotent algebraic group $U$ to representations of its overgroup $U_n$, then I get the theorem for free for $U$. Of course, I have no reason to believe when and if this is true, it’s just a shot in the dark that gives me a wonderful shortcut.

Ideas?