How complex is the orbit equivalence relation of $\mathrm{Iso}_0(X)\curvearrowright S_X$ for $X=L^p([0,1])$? For a Banach space $X$ let $S_X$ denote its unit sphere and let $\mathrm{Iso}_0(X)$ denote the group of rotations of $X$, that is isometries fixing the origin. There is a natural continuous action $\mathrm{Iso}_0(X)\curvearrowright S_X$.
When $X=L^p([0,1])$ the group $\mathrm{Iso}_0(X)$ is Polish, so we can ask how complex the orbit equivalence relation induced on $S_X$ is in the descriptive set theoretic sense. By a result of Pełczyński and Rolewicz every orbit is dense, but I don´t know much apart from this.
Are orbits meagre? Does this equivalence relation admit classification by countable structures? What are good references for this kind of questions and for more general questions along those lines (for example when $X$ is the Gurarij space)?
 A: The orbit structure is extremely simple. If $p=2$, there is one orbit (the isometry group of a Hilbert space acts transitively), whereas for $p\neq 2$ there are exactly $2$ orbits: the (classes of) functions that do not vanish on a set of positive measure and its complement, the functions that do vanish on a set of positive measure.
This is certainly well-known, but I do not remember where this is written (I thought that it might in a paper by Ferenczi and Rosendal, but I could not locate it). Let me provide the proof in the interesting case $p\neq 2$.
Let me first prove that if two functions $f$ and $g$ in $S_X$ do not vanish, then they are in the same orbit. The two probability spaces $([0,1], |f|^p d\lambda)$ and $([0,1], |g|^p d\lambda)$ (where $\lambda$ is the Lebesgue measure) are standard complete atomless probability spaces. They are therefore isomorphic. This implies that there is a bimeasurable bijection $\phi \colon [0,1] \to [0,1]$ which sends $|f|^p d\lambda$ to $|g|^p d\lambda$. The three maps

*

*$h \in L^p(d\lambda) \mapsto \frac{h}{f} \in L^p(|f|^p d\lambda)$,

*$h \in L^p(|f|^p d\lambda) \mapsto h \circ \phi^{-1} \in L^p(|g|^p d\lambda)$,

*$h \in L^p(|g|^p d\lambda) \mapsto gh \in L^p(d\lambda)$.

are all surjective isometries (here we use that $|f|>0$ and $|g|>0$ a.s.). Their composition $h \mapsto g \frac{h \circ \phi^{-1}}{f \circ \phi^{-1}}$ is therefore a linear isometry which maps $f$ to $g$.
If $f$ and $g$ both vanish (say on  $A$ and $B$ respectively), we can find a linear isometry from $L^p(A,d\lambda)$ onto $L^p(B,d\lambda)$ and combine it with a linear isometry $L^p([0,1]\setminus A) \to L^p([0,1]\setminus B)$ given by the previous case to find a isometry of $L^p([0,1])$ that maps $f$ to $g$.
The converse (that a function that almost surely does not vanish cannot be in the orbit of a function that vanishes) follows from the Banach-Lamperti theorem, which characterizes the linear isometries of $L^p([0,1])$: all such isometries are of the form
$$ Uf(x) =\omega(x) f(T^{-1}(x))$$
for a measurable bijection $T \colon [0,1] \to [0,1]$ that preserves the class of the Lebesgue measure $\lambda$, and a function $\omega\colon[0,1] \to \mathbf{C}^*$ satisfying $|\omega(x)|^p = \frac{d T_*\lambda}{d\lambda}(x)$.
