Is Hamiltonian evolution unique given a fixed input and output state? Consider a quantum system that is prepared at some initial state $| \text{in}\rangle$ and we are able to measure the output state $| \text{out}\rangle$. Furthermore assume this evolution of the state is governed by a Hamiltonian without any control signals and let us consider for simplicity that the system is closed (such simulations can be easily run on Qutip).
Essentially:
$$
H:| \text{in}\rangle \mapsto | \text{out}\rangle 
$$
This can easily be visualized by considering a qubit. In the Bloch sphere this evolution corresponds to a curve whose initial point is the input state and the final point is the output state.
Question:
Given fixed $|\text{in}\rangle,|\text{out}\rangle$ is there a unique Hamiltonian that maps $|\text{in}\rangle \to |\text{out}\rangle$?
Thinking of paths, we know that there are many inequivalent paths from one state to another. Nevertheless, nature will choose the one that minimizes the path integral. This amounts into minimizing the Lagrangian. Furthermore, thinking of paths again, one can find symmetries on the system which are redundant.
Thinking of Hamiltonians, is there something equivalent? Specifically:

*

*Is it known what are the symmetries in the space of Hamiltonians that results in equivalent evolution $|\text{in}\rangle \to|\text{out}\rangle$?

Naively thinking, I can think of some toric action (a phase if you will) multiplying any Hamiltonian that should produce the same class of evolution and that can be normalized (choosing phase to be equal to unity).
Layman's terms:
Is the evolution $|\text{in}\rangle \to |\text{out}\rangle$ be given by a unique Hamiltonian and what is actually known about it regardless of the answer?
 A: This is too long for a comment but consider the following situation. Suppose $\vert \uparrow\rangle$ and $\vert\downarrow\rangle$ are you initial and final states.
Then $e^{i\pi \sigma_y/2}\vert\uparrow\rangle=\vert \downarrow\rangle$ while $e^{i\pi \sigma_x/2}\vert\uparrow\rangle =i\vert\downarrow\rangle$, so you get the same final state up to a phase.
If you allow for piecewise Hamiltonian, then you can always rotate away the phase $i$ in $i\vert\downarrow\rangle$ using $e^{i\pi\sigma_z/2}$.
If you do allow final states up to a phase, there are clearly more solutions since you can go from the North Pole to the South Pole of the Bloch sphere along any line of constant latitude.
If you do allow piecewise Hamiltonian, you can find even more solutions.  Pictorially, you can imagine going from the North Pole to the equator, travel some distance along the equator, and then from the equator to the South Pole.
You might have to include piecewise Hamiltonians if you have constraints on your Hamiltonians: there are some groups which are not simply connected so you might more than one transformation to reach a final state from an initial state.  An example would be an $\mathfrak{su}(1,1)$ evolution, where the exponential map does not cover the whole group with a single exponential.  Details can be found in

Chiribella G, D’Ariano GM, Perinotti P. Applications of the group SU (1, 1) for quantum computation and tomography. Laser physics. 2006 Nov;16(11):1572-81.

or the arXiv version.
