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?


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