The two sequences A48680 and A48679 of the OEIS define two mutually inverse bijections on the set of all strictly positive natural numbers given (for the comfort of the reader) as follows:
Given an integer $n>0$ with binary expansion $n=\sum_{i=0}^N d_i2^i$, make the substitution $1\longrightarrow 10,\quad 0\longrightarrow 0$ in $d_Nd_{N-1}\ldots d_0$, remove the final digit $0$ and interpret the resulting word (whose digits $1$ are isolated) $e_Me_{M-1}\ldots e_1$ as the Zeckendorff expansion of $\varphi(n)=\sum_{i=1}^M e_if_i$ (where $f_0=f_1=1,f_n=f_{n-1}+f_{n-2}$ is the Fibonacci sequence).
Example: For $n=13=8+4+1$ we get $$1101\longrightarrow 101001(0)\longrightarrow f_6+f_4+f_1=13+5+1=19$$ showing $\varphi(13)=19$.
It is easy to see that $\varphi$ defines a bijection which has a few small finite orbits: $$(1),(2),(3,4),(5,6,7,12,11,17,14,20,16,8),(9),(10)\ .$$ (I suspect that this is the complete list of finite orbits).
The orbit of a "generic" integer should be infinite: Indeed most binary integers with a large number $N$ of digits have roughly N/2 digits equal to 1 suggesting that the number of digits of under iteration should have roughly geometric growth of ratio $\frac{3\log(\omega)}{2\log(2)}$ for $\omega=\frac{1+\sqrt{5}}{2}$ the golden number (where we make of course the perhaps naive assumption that digits $0$ and $1$ are occuring with roughly identical frequencies during iterations).
This heuristic argument does however break down when considering backward iterations.
A few computations of $n\longmapsto \log\log(\varphi^n(a))$ for various initial values $a$ (with infinite orbits) and $n$ in the range $-200,\ldots,200$ depict however graphs which seem to be (up to affine transformations) asymptotically close to the graph $x\longmapsto \vert x\vert$ (with observed asymptotic slopes seemingly fairly close to the expected values $\pm \log\left(\frac{3\log(\omega)}{2\log(2)}\right)$) suggesting that the same (or roughly the same) asymptotical geometric growth-rate occurs also for digits of iterated values when considering the reciprocal function.
Is this an artefact due to numerical limitations? If not, are there some easy (perhaps heuristic and not completely rigourous) explanations for these behaviours.
Final Remark: The definition of $\varphi$ mixes two different numeral systems. It is thus somewhat reminiscent of the famous Syracuse problem.
