What does it mean for a system to be invariant under rotation? This question is a query to clear up confusion regarding invariance under rotation and the associated conservation of angular momentum.
The confusion arose while studying the following problem:

A particle with the mass $m$ is moving without friction on a plane surface from A to B with the velocity $v_0$. After the particle passes the point B, someone starts pulling in the particle using a thin string originating from the point P. The particle then moves in a half-circle from B to C. When the particle reaches the point D, the pull from the string stops and the particle moves towards D with constant velocity.
The point O is the midpoint between B and C, and the center of the half-circle from B to C. The point P is the midpoint between O and C.

  
  
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*What is the velocity of the particle at the point C?
  
*How much work does the string do on the particle, when it moves from B to C?
  
*How long does it take to move from B to C.
  

I was studying the solution of the problem with a friend. The solution involves using conservation of angular momentum, and it’s pretty obvious that it’s conserved around P, since the only force has no tangential parts compared to P.
While talking about this conservation of angular momentum, my friend mentioned something called Noether’s theorem, and said that angular momentum is preserved since the system is invariant under rotation around P.
It makes sense enough that the system doesn’t change under rotation around P – everything is simply rotated, and so what? I don’t get why that implies conservation of angular momentum, and I don’t see why it’s not rotation-invariant around another point like B. It looks like I’ve misunderstood what it means to be rotation-invariant.
A complete answer to this question would answer the following: What does it mean to be rotation-invariant, and why does it imply conservation of momentum? Why is the listed problem rotation-invariant around P, but not B?
I have not studied Lagrangian or Hamiltonian mechanics.
 A: Noether's theorem states that for every conserved quantity in a dynamical system there is a symmetry associated and vice versa. 
For example, suppose you have a system for which the energy is conserved. Then whether you perform an experiment with it right now or whether you perform the experiment $\delta t$ time later, you will get the same results. The system is invariant under time translation.
Suppose, you have a system for which momentum is conserved in a particular direction, say, the collision of two billiard balls. Then whether you displace the system in the same direction and perform the experiment or you perform it here you will get the same results. Contrast this with the system on a sloping billiards table. For that, just as you displace the system the momentum of the balls would increase (due to gravity).
Coming to your question, rotation-invariant about a point means invariance in the results when performing the experiment as it is and when performing it after it has been (instantaneously) rotated by any arbitrary angle about that point. For example, suppose I am on the surface of the earth, in London say, and I perform some experiment (say swinging a pendulum). Now, suppose I instantaneously rotate myself about the earth and reach Paris. Now provided there have been no altitude changes and the rotation was instantaneous (because we are not assuming energy conservation) I would get the same results for the experiment.
Finally, it is not invariant about point B because suppose you rotate the experiment about B such that the point P passes through the initial trajectory of the particle. Then the resulting motion would not be the same. As I think of it, rotating the Hamiltonian is analogous to rotating the configuration of the setup - the potentials and the sources of forces, etc. - and not the initial conditions of the trajectory.
