It is fair to assume it will follow a great circle. This path is a geodesic on a 2-sphere - the object experiences no forces, angular momentum is conserved. To determine positions, why not use the equations $\theta(t)=\theta_0+\dot\theta t$ for both polar and azimuthal angles?

I don't think is the best prediction. I argue the best prediction is the one that would minimize the angular momentum of the system. And that is not geodesic rotation, but one in between [θ,Pi]

you also need to know how many turns the particle has completed between the two measurements

Let's explore the full turns < 1 case, all others would imply a larger angular momentum, making it less likely.

I get it that in many systems, we find the rotation to be close to the greatest circle, say planets around a star. But in Quantum Physics, it doesn't have to be. What precludes the particle from escaping that higher latitude orbit? It doesn't matter. This thought exercise aims to explore the application of Principle of Least Action on rotation.

@josephh how would you choose the azimuthal angle for your prediction?

Only great circles are geodesic, so if you permit small circle paths there must be some force keeping the particle on the circle. Also, 2 points aren't sufficient to determine a circle (either on the Euclidean plane or on a sphere), so there will be an infinite family of solutions.

how would you choose the azimuthal angle for your prediction? What does that mean?

I don't think is the best prediction. I argue the best prediction is the one that would minimize the angular momentum of the system. No. And that is not geodesic rotation, but one in between [θ,Pi] I'm not sure what that means either. Are you saying the object experiences forces? You did not mention that in your OP.

I'm surprised at the community’s unwillingness to consider latitudinal orbits on this question. Imagine if needed another particle in a symmetrical latitudinal orbit on the other side respect the origin or rotation plane. Physicists have fallen into the shortest distance trap before: Fermat’s Principle/Snell’s Law

@josephh In the video I show that indeed the two observations can be explained by an infinite number of possible rotations if we consider rotating the point over some axis. If you calculate the angular momentum for each, you'll find a clear minimum. I argue this is the most likely, absent of any other information, as it minimizes the needed system energy.

"Physicists have fallen into the shortest distance trap before: Fermat’s Principle/Snell’s Law" Okay, if you have a better idea, then write a paper which falsifies this principle. It will get you a Nobel Prize.

"if we consider rotating the point over some axis." Is the particle free to move or not? The question and your comments contain multiple contradicting statements. That is why there's "unwillingness to consider latitudinal orbits on this question", which is also unclear. You need to rewrite your question so that it makes sense if you want an answer.

The problem is underdefined. Energy is conserved if your lagrangian is time-independent. If we do not know the energy we at least need to know its distribution to make any claims about the system. Also, we need to know if there is any potential or a constraint that would keep the particle from following the great circle. You cannot use the least action principle to discriminate between systems with different parameters.

yes, if you have extra forces the particle can move along any circle. In such a case you need at least three points, as there is an infinite number of circles on the sphere that include two specific points

@Wolphramjonny Yes, there is an infinite number of circles on the sphere that include two specific points. So if that's all I know, to decide where should I point my tele/micro/scope to, what should I choose? The geodesic, the shortest circle, or one in between? How about the one with the least angular momentum?

@MartinVítVavřík Yes, "we at least need to know its distribution to make any claims about the system". In this problem, that distribution is quite knowable, as you can assume a few things without loss of generality.

There is geometric beauty to minimizing angular momentum on that distribution

No way to tell with just two points it could be any!

The distribution is not "quite knowable" if we have no information about the system. And if you presuppose that the minimal energy is the most likely, then clearly the shortest path between the two points (which is along the great circle) is the most likely (if there is no potential and movement along any circle can be enforced by an arbitrary constraint). The kinetic energy in this system is one half of the product of the angular momentum and the angular velocity, so the trajectory minimizing angular momentum isn't the trajectory minimizing energy.

you can chose any extra condition you want, but it will be an arbitrary choice