Assign $x'$ to a stone in the $i$th glass. Let's consider what happens to the sum of the number assigned to all stones with each movement.

Movement A replaces $x^i + x^{i+1}$ with $x^{i-1}$; movement B replaces $2x^i$ with $x^{i-1} + x^{i+2}$. So the difference in the sum is either $x^{i-1}(1 - x - x^2)$ or $x^{i-1}(1 + x^3 - 2x) = x^{i-1}(1 - x - x^2)(1 - x)$.

Choose $x$ so that $1-x-x^2=0$, say $x = \phi^{-1} = \frac{\sqrt{5}-1}{2}$. Then the sum $S$ of the numbers assigned to the stones always stays the same. There is a number $x^k$ such that $S < x^k$, so there is a limit to the leftmost stone, so it won't move after a while. Now consider the sum without this stone; the leftmost stone (that is, the second leftmost stone overall) won't move after a while and so on. So the configuration will remain constant at some time, that is, there won't be any two stones in the same glass nor two stones in two consecutive glasses.

Now notice that the representation of $S$ in base $\phi^{-1}$ (all digits equal to one, no two consecutive ones) is unique: suppose on the contrary that there are two such representations and let $k$ be the first position from left to right in which those representations differ. This means that $\phi^{-k}$ is equal to the sum of some numbers of the form $\phi^{-m}$ for $m > k + 1$. But $\sum_{m=k+2}^{\infty} \phi^{-m} = \frac{\phi^{-k-2}}{1-\phi^{-1}} = \phi^{-k}$, so this isn't possible. So the final configuration is unique.