Suppose I have a morphism $f:X\to Y$ which is a GIT quotient of $X$ with respect to some reductive, linear group. 
Does the semistable $X^{ss}$ and stable locus $X^s\subset X$ determine completely the linearization (maybe up to taking a power of the linearization itself)?


Or, in better words, can two different linearizations $L$ and $L'$ of different GIT quotients of $X$ give the same semi-stable and stable locus?



Does the semi-stable set determine the linearization of a GIT quotient?
The paper you want to have a look at is Thaddeus, [GIT and Flips](https://doi.org/10.1090/S0894-0347-96-00204-4 "J. Am. Math. Soc. 9, No. 3, 691-723 (1996), arXiv:alg-geom/9405004. zbMATH review at https://zbmath.org/0874.14042") ([JSTOR](https://www.jstor.org/stable/2152810)), or Dolgachev–Hu, [Variation of GIT quotients](https://doi.org/10.1007/BF02698859 "Publ. Math., Inst. Hautes Étud. Sci. 87, 5-56 (1998). zbMATH review at https://zbmath.org/1001.14018") ([Numdam](http://www.numdam.org/item?id=PMIHES_1998__87__5_0)).


There are finitely many polyhedral chambers within the space of possible linearizations. In the interior of each chamber, the sets $X^s$ and $X^{ss}$ are constant. Crossing a wall, these sets change; in nice situations, the two GIT quotient $X^{ss}/G$ on each side of the wall are related by a flip.


