Let $f:G\to H$ be a surjective morphism of Lie groups. Suppose $g\in G$ is a unipotent element, namely $Ad_g:Lie(G) \to Lie(G)$ is a unipotent transformation of Lie algebras(the only eigenvalue of is one). Now I wonder how to show that $df(g)\in H$ is also unipotent.
My attempt: Let $w\in Lie(H)$ be a eigenvector of $Ad_{f(g)}$, say $Ad_{f(g)}(w)=rw$. Since $f$ is an epimorphism of Lie groups, I remember that $df$ is surjective as well and assume $df(v)=w$. So we have
$$Ad_{f(g)}(df(v))=r df(v).$$
I wish to somehow conclude from here that
$$Ad_g(v)=rv$$
and since $g$ is unipotnent we have $r=1$. But how? Or are there any other approaches.
For any Lie group morphism $f\colon G\to H$ the conjugation map $C_g\colon h\mapsto ghg^{-1}$ satisfies the identity $$f\circ C_g = C_{f(g)}\circ f.$$ Differentiating this leads to a similar formula for the adjoint map \begin{equation}\label{1}\tag{*} df\circ\mathrm{Ad}_{g} = \mathrm{Ad}_{f(g)}\circ df. \end{equation} Hence if $X$ is an eigenvector of $\mathrm{Ad}_{g}$ and $df(X)$ is nonzero, then $df(X)$ is an eigenvector of $\mathrm{Ad}_{f(g)}$ with the same eigenvalue.
The same is true also for generalized eigenvectors: Suppose $(\mathrm{Ad}_{g}-\lambda\cdot \operatorname{id})^nX=0$ for some $n\in\mathbb{N}$ and eigenvalue $\lambda$. The binomial expansion of the operator $(\mathrm{Ad}_{g}-\lambda\cdot \operatorname{id})^n$ is $$(\mathrm{Ad}_{g}-\lambda\cdot \operatorname{id})^n = \sum_{k=0}^n(-\lambda)^k\binom{n}{k}\mathrm{Ad}_g^k.$$ Repeated application of formula \eqref{1} then shows that $$(\mathrm{Ad}_{f(g)}-\lambda\cdot \operatorname{id})^n(df(X)) = df\Big((\mathrm{Ad}_{g}-\lambda\cdot \operatorname{id})^nX\Big) = df(0) = 0,$$ so if $df(X)$ is nonzero, then it is a generalized eigenvector of $\mathrm{Ad}_{f(g)}$ with the same eigenvalue $\lambda$.
Decomposing the Lie algebra $\mathrm{Lie}(G)$ as a direct sum of the generalized eigenspaces of the map $\mathrm{Ad}_g$ then shows that the spectrum of $\mathrm{Ad}_{f(g)}$ restricted to the image of $df$ is a subset of the spectrum of $\mathrm{Ad}_g$. In particular, if $f$ is surjective and $\mathrm{Ad}_g$ unipotent, then $\mathrm{Ad}_{f(g)}$ is also unipotent.