Let $G\to \mathbb A^n_{\mathbb C}$ be a smooth morphism whose fibers at any point of $\mathbb A^n$ are unipotent groups. Can we conclude that $G\simeq \mathbb A^{n+N}_{\mathbb C}$ for some $N$, as a scheme? This would be true for $n=0$ by a theorem of Lazard, but it's not clear to me whether it can be generalized.
If (as I suspect) this is not true, maybe there are some more hypotheses one can put?
Related: can one deduce from the given conditions that $G\to \mathbb A^n_{\mathbb C}$ is a torsor?
The main example I have in mind is the Beilinson-Drinfeld version of the kernel $G_{\mathcal O}^{(j)}\to G_{\mathcal O}^{(j')}$, where $j\geq j'$, $G$ is a reductive complex group, and $G_{\mathcal O}^{(j)}$ is the truncation of the arc group at the $j$-th power. We know this kernel is unipotent, hence isomorphic to $\mathbb A^N$ for some $N$, and I wonder whether an analogous result is true in the Beilinson-Drinfeld setting.
