$ \renewcommand{\C}{{\mathbb C}} \renewcommand{\R}{{\mathbb R}} $ In the preprint Taking quotient by a unipotent group induces a homotopy equivalence we proved the following result:
Theorem. Let $U$ be a unipotent group over $\C$ acting on an irreducible complex algebraic variety $X$. Assume that there exists a surjective morphism of complex algebraic varieties $\varphi\colon X\to Y$ whose fibres are orbits of $U$. If $X$ and $Y$ are smooth and all orbits of $U$ in $X$ have the same dimension, then the induced map on the $C^\infty$-manifolds of $\C$-points $X(\C)\to Y(\C)$ is a homotopy equivalence. Moreover, if $U$, $X$, $Y$, and $\varphi$ are defined over $\R$, then the induced map on the $C^\infty$-manifolds of $\R$-points $X(\R)\to Y(\R)$ is surjective and induces homotopy equivalences on connected components.
Note that the assumption that all orbits of $U$ in $X$ have the same dimension is equivalent to the assumption that the morphism $\varphi$ is smooth; see Ravi Vakil, "The Rising Sea. Foundations of Algebraic Geometry", Theorem 25.2.2 and Exercise 25.2.F(a).
The only examples for this theorem that I know are the following:
Trivial example 1. $X=G$ is a connected algebraic $k$-group (where $k=\C$ or $k=\R$), $\ U\subset G$ is a unipotent $k$-subgroup, $Y=G/U$, $\ \varphi\colon G\to G/U$ is the quotient morphism.
Trivial example 2. $G$ is a connected algebraic $k$-group (where $k=\C$ or $k=\R$), $U'\subset G$ is a unipotent $k$-subgroup, $N\subset U'$ is a normal $k$-subgroup of $U'$, $\ X=G/N$, $\ Y=G/U'=X/U$ where $U=U'/N$, $\varphi\colon G/N\to G/U'=X/U$ is the quotient morphism.
Question. What is a nontrivial example of such $(U,X,Y,\varphi)$ ?
