Based on Eran Zimmerman Gonen answer:
what about this flow graph:
G=(V,E), s.t V={s,a,b,c,t}, E={(s,a),(a,b),(b,c),(c,t)}.
the capacity of all edges is 1, so as the flow.
s-(1/1)->a-(1/1)->b-(1/1)->c-(1/1)->t
So, based on the residual graph, I have a min-cut S={s}.
Starting from t I am getting S'={t,c,d,a}.
Looking at 7th label, V/S' = {s} = S which means I have only one min-cut, but I have many more: S={s}, S'={s,a}, S''={s,a,b}, S''' = {s,a,b,c}