Let $E$ be a finite real dimensional vector space and $\{u_t\}_{t\in \mathbb R}$ be a unipotnent one-parameter subgroup of $GL(E)$, which means each $u_t$ is unipotent (the only eigenvalue is $1$) and $u_{s+t}=u_s \circ u_t$.
I wonder how to derive from the above information that $u_t$ must be of the form
$$u_t=\exp(t \eta),$$
where $\eta$ is a nilpotent transformation on $E$.
The key part is to find $\eta$ I guess.
Write $A={d\over{dt}}_{t=0}u_t$, consider $f_t=exp(tA)$, $f_t$ and $u_t$ are solutions of the same OD and coincide at $0$, so they coincide. This implies in particular that $exp(A$ is unipotent and $A$ is nilpotent, otherwise $A$ has an eigenvalue eventually complex distinct of $0$.