Non-recurrent points of $F(a,b)=(b,ba)$ in a compact metric group $G$ Consider a compact metric group $G$ [A compact topological group $G$ where the topology is generated by an invariant metric].  I am particularly interested in the case where $G$ is the $n$-dimensional projective unitary group $\mathrm{PU}(n)$, but the following is phrased more generally.
Consider the following "Fibonacci" map
$$F:G\times G\to G\times G\\
  \quad\quad(a,b)\mapsto (b,ba).$$
[The name comes from the fact that, when iterated, $F$ produces the Fibonacci words on $G$]
It is easy to see that $F$ is continous and the Haar measure on $G\times G$ stays invariant under the application of $F$. The latter follows from Fubini's theorem and the invariance of the Haar measure.
We say that $(a_0,b_0)\in G\times G$ is $F$-recurrent if there exists a subsequence of $(F^n(a_0,b_0))_{n\in \mathbb{N}}$ that converges to $(a_0,b_0)$. Let $R\subseteq G\times G$ be the set of all $F$-recurrent points. By the Poincaré recurrence theorem, the set $N= G\times G\setminus R$ has zero Haar measure.
I am trying to understand the form of the (possible?) non-$F$-recurrent points $(a_0,b_0)\in N$, or, even better, I would like to prove that $N=\emptyset$, which I suspect is the case. Is there a stronger version of the Poincaré recurrence theorem that would allow me to conclude that $N=\emptyset$ for this map?
I have looked around for general references that study this specific map but have not found much. This is a bit far from my usual research area so I might be lacking some keywords. Ideas are welcome.
Update:
As commented, $N\neq \emptyset$. Is there a simple condition on $(a,b)\in G\times G$ that guarantees $(a,b)\in R$?
For example, is the following true?

*

*If the subgroup generated by $\{a,b\}$ is dense in $G$, then $(a,b)\in R$?

 A: Here's a very partial answer. Let $R_G\subset G^2$ be the set of recurrent points of $F$.

Then for a compact group $G$, we have $R_G=G^2$ if and only if $G$ is profinite.

Proof. First, for $G$ the circle group $\mathbf{R}/\mathbf{Z}$, $R_G\neq G$. Indeed the map $F$ is then given on the 2-torus $(\mathbf{R}/\mathbf{Z})^2$ by the matrix $\begin{pmatrix}0&1\\1&1\end{pmatrix}$, which has the eigenvalue $s=-(\sqrt{5}-1)/2$. If $(a,b)\neq (0,0)$ is any pair in the corresponding eigenline (which is dense in the torus), then $F^n(a,b)$ converges to (0,0), so $(a,b)$ is not recurrent.
In general, observe that the condition $R_G=G^2$ passes to both closed subgroups and to quotients. If $G$ is a compact non-profinite group, then by Peter-Weyl it has a quotient that is a positive-dimensional compact Lie group, which in turn has a closed subgroup isomorphic to the circle group.
Conversely, suppose that $G$ is profinite. For $(a,b)\in G$, let $N$ be a normal open subgroup. Then $F$ induces a permutation of $(G/N)^2$. So for some $n_0$, $F^{n_0}$ induces the identity on $(G/N)^2$. Hence for every $n$, $F^{nn_0}(a,b)\in (a,b)N^2$. Hence $(a,b)$ is recurrent. (Note that this even shows that $F^n$ converges uniformly to the identity when $n$ tends to 0 in the profinite completion of $\mathbf{Z}$. Concretely, for instance, this shows that $F^{n!}$ converges uniformly to the identity.
