My question is: Are the orbits of the geodesic flow on $S^n$ determined as the fibers of the momentum map for its $SO(n+1)$ symmetry?

I started by considering the analog problem for the orbits of the hamiltonian flow of the n-dimensional harmonic oscillator and the fibers of the momentum map for its $U(n)$-symmetry. If I am not wrong in such a case the answer is yes.

Below I give some other details, hoping to be sufficiently clear.

*Any kind of correction and/or suggestion is really welcome.*

**The n-dimensional harmonic oscillator**
Let us consider the n-dimensional isotropic harmonic oscillator as the hamiltonian system on $\mathbb{C}^n$ with the symplectic form $\omega=\sum_{k=1}^n d\overline{z}^k\wedge dz^k$ and the Hamilton function $H(z)=1/2|z|^2$.

The natural action of $U(n)$ leaves $H$ invariant and has an equivariant momentum map given by $\langle J,A\rangle(z)=1/2\sqrt{-1}\langle z,Az\rangle$ for $z\in\mathbb{C}^n,\ A\in\mathfrak{u}(n)$.

The fibers of $J$ are exactly the orbits of the hamiltonian flow, infact $J^{-1}(J(z))=S^1.z\equiv\{e^{i\phi}z|\phi\in\mathbb{R}\}$.

**The geodesic flow on $S^n$**
We can consider the geodesic flow on $S^n$ as the hamiltonian system on $TS^n\equiv\{(x,y)\in T\mathbb{R}^{n+1}\equiv\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}||x|=1,\langle x,y\rangle=0\}$ with the symplectic form induced on it by the canonical symplectic form $\sum_{k=1}^{n+1} dy^k\wedge dx^k$ on $T\mathbb{R}^{n+1}\equiv\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}$, and the Hamilton function $H(x,y)=\frac{1}{2}|y|^2$.

$TS^n$ is invariant under the lifting to $T\mathbb{R}^{n+1}$ of the natural action of $SO(n+1)$ on $\mathbb{R}^{n+1}$.

This action leaves $H$ invariant and has an equivariant momentum map given by $\langle J,A\rangle(x,y)=\frac{1}{2}(\langle x,Ay\rangle-\langle y,Ax\rangle)$ for $(x,y)\in TS^n,\ A\in\mathfrak{so}(n+1)$.

On $T^{\times}S^n\equiv(TS^n)\setminus S^n$, the punctured tangent space to $S^n$, the momentum map has constant rank $2n-1$. So, for any $(x,y)\in T^{\times}S^n$, we have that $J^{-1}(J(x,y))$ is a $1$-dimensional submanifold and includes $\{(e^{t.J(x,y)}x,e^{t.J(x,y)}y)|t\in\mathbb{R}\}$, the orbit of the hamiltonian flow through $(x,y)$, but I don't know if this is its only component.

My question amounts to: Is $J^{-1}(J(x,y))$ exactly equal to $\{(e^{t.J(x,y)}x,e^{t.J(x,y)}y)|t\in\mathbb{R}\}$? or not? Where $J:T^{\times}S^n\to\mathbb{so}(n+1)^\ast\cong\mathfrak{so}(n+1)$ is given by $J(x,y)=\frac{1}{2}(y^Tx-x^Ty)$, (with the linear isomorphism $\mathbb{so}(n+1)^\ast\cong\mathfrak{so}(n+1)$ realized through the scalar product trace).

**Edit**
Now I have realized that there is an easy positive answer to my question and I have posted it below. Again any kind of comment is welcome.