Well, we assume that the robot moves, not the map. Map stays the same all the time, it's just cyclic. Imagine that the robot is moving around a round stadium with several entrances. To map it, we declare that one place is a start, but since the "building" itself is round, once you reach the far right end on map, you actually end up at the start. But still, it's you (or the robot!) that moves, not the building.

 *
 1 2 3 4 5 
 |_______|  move right

   *
 1 2 3 4 5
 |_______|  move right

     *
 1 2 3 4 5
 |_______|  move right

       *
 1 2 3 4 5
 |_______|  move right

         *
 1 2 3 4 5
 |_______|  move right

 *
 1 2 3 4 5
 |_______|  move right

It's the robot that moves... If you remember the location vs probability graph that was drawn at the beginning of the localization lecture, that "location" is actually the "robot's location in the world" and the probability is the "probability of the robot's belief that it's at the correct location". The "bumps" on that graph will move as the robot moves. Ideally, as the robot moves the probability of it being correctly localized will increase and "travel" (shift cells) with the robot as it moves left/right or up/down. Eventually, it will find itself in the cell with the highest probability (biggest bump on the loc. vs prob. graph) therefore successfully localizing itself in the given "world".

Moving the robot to the right is the same as moving the world to the left. And since you are trying to localize the robot you don't know exactly where it is, so you don't have a robot marker to move around. But you do have a world (believe probability of the world).

A - B - C - D - E    
1 - 2 - 3 - 4 - 5

Say the robot was at position C observing 3 and you do a right move. Then the robot should be at a position observing 4. After moving the believe world left (to simulate the right move).

A - B - C - D - E
2 - 3 - 4 - 5 - 1

And this will work for when the robot was at A, B, D or E too.

This is how I understand it. If it moves to the right exactly one, it still has the same probability that it knows where it is. In other words, it knows it moved exactly one space and so the probability of where it is at now is no better or worse than before. That is why the probability distribution shifts to the right with the robot.

No. The world should be

1 - 2 - 3 - 4 - 5
    *

Whence, what you believed to be true when you were on 1 is now believed to be true in 2. The same follows for cell.

You're absolutely right if you view it as shifting the world instead of shifting the probability vector. Since you've shifted the world, you can look up the probabilities in the original vector. So the probabilities for 2 to 5 and then 1 are now: (0.11, 0.33,0.33,0.11,0.11). Since you're asked to give them for 1 to 5, just copy down the last followed by the first four:(0.11,0.11,0.33,0.33,0.11) which is the right answer.

I believe Prof Thrun mentions that the motion in the example is moving two steps to the right. In this case the robot moves from cell 1 to cell 3 (given exact motion). This was probably in order to be able to better show undershoot (which would be moving one step to the right and not staying in the same place).