I think I have found an example that shows that the answer is YES. Two differente linearizations may give the same (semi-)stable locuses. I would be glad if someone could prove me wrong, though!

Thank you for your comments, the statement has been edited and now it looks much better!

@ steven: yes, of course!

@ulrich: that's a good counterexample, thank you. I guess I have to assume $Y$ and $Z$ normal (or maybe just one of them) In that case I believe that the statement should be true. @Laurent: yes, you are right, thank you. basically that (very weak) assumption descends from the other hypothesis. I will edit the question.

In fact I mean exactly that (I did some editing adding an irreducibility hypothesis, thank you!): are the hypothesis enough to assure that $Y=Z$ and the morphism is the same. I mean that $Y$ and $Z$ have the same underlying set, no topology.

I mean that there's a one-to-one map between the set of closed points of $Y$ and the set of closed points of $Z$.