I am reading https://www.sciencedirect.com/science/article/pii/S138572587680008X and must be missing something obvious. The premise is that we are considering the fixed point set of a unipotent transformation $u$ on the flag variety.
Now I am assuming that the action is by right multiplication, in the same way as we can identify the stabilizer of the entire flag variety with the upper triangular matrices. So if $u$ fixes any flag, looking at each column in succession...how can it not be an upper triangular matrix? In which case it would fix all the flags.
Note that the flag variety is $G/B$, hence consists of the left cosets $gB$ for $g\in G$. With this definition, it does not make sense for a matrix to act by right multiplication. We would end up with $gBh$, which could lie in multiple different left cosets. If we were considering the right cosets, which is obviously completely equivalent, then the action would be by right multiplication.
Even the stabilizer of a fixed flag, the Borel subgroup itself, does not leave all points fixed in the variety with the left action. Indeed, the double cosets $BgB$ are extremely important. They depend only on an element $w\in N(T)/T$, the normalizer of the maximal torus modulo the torus itself, which is the Weyl group, and the double cosets $BwB$ are the Schubert cells corresponding to $w$ in the Weyl group.