# If $\phi : G \to G$ is a Lie group automorphism without fixed points, is $g^{-1}\phi(g)$ surjective?

If $$G$$ is a real connected Lie group and $$\phi : G \to G$$ is a Lie group automorphism without fixed points other than $$e$$, does it follow that the map $$\psi : g \mapsto g^{-1} \phi(g)$$ surjective? Can we describe its inverse?

Note that that $$\psi$$ is injective. (Automorphism with no fixed points other than identity) The example $$G = \mathbb Z$$ and $$\phi(n) = -n$$ shows that we need $$G$$ connected.

Context: I was reading a computation in Iwaniec's Spectral Methods of Automorphic Forms, where the case of $$G = N = \left\{\begin{pmatrix}1 & * \\ 0 & 1\end{pmatrix} \right\} \subset \mathrm{SL}_2(\mathbb R)$$ and $$\phi = \mathrm{Ad}_a$$ with $$a = \begin{pmatrix}y & 0 \\ 0 & y^{-1}\end{pmatrix}$$, $$y > 1$$, occurs in a computation for the Selberg trace formula. The inverse of $$\psi$$ is easily computed as $$\begin{pmatrix}1 & n \\ 0 & 1\end{pmatrix} \mapsto \begin{pmatrix}1 & n/(y^2 - 1) \\ 0 & 1\end{pmatrix}$$, but I want to understand this conceptually.

Idea: in the example above we have, by Taylor expansion, $$\psi^{-1}(n) = \prod_{k = 1}^\infty \mathrm{Ad}_{a^{-k}}(n)$$ Maybe in the general case, $$\psi^{-1}(g) = \cdots \phi^{-3}(g) \phi^{-2}(g) \phi^{-1}(g)$$ ? Provided the product converges?