When $\frac{\text{Aut}(G/P,L)}{S^1}$ is discrete? Let $(M,\omega)$ be a Kähler manifold with a pre-quantum Line bundle $L$ and
  $\text {Aut}(M,L)$ means the group biholomorphisms of $M$ which lift to holomorphic bundles maps $L\to L$. My question is for the flag variety $G/P$ when $\frac{\text{Aut}(G/P,L)}{S^1}$ is discrete? 
PS; I think $\text{Aut}(G/P,L)$ is equal to $\text{Aut}(G/P)$ but I am not sure.
 A: Without knowing anything in particular about the $S^1$-action, your condition seems to me unlikely to be satisfied very often. Let $G=SL_2(\mathbb{C})$ and let $P$ be the standard Borel of upper-triangular matrices. Given any weight $n\in\mathbb{Z}$ of the standard maximal torus $T$, we can construct an associated line bundle $$L=\mathcal{O}(n):=G\times_{P}\mathbb{C}(n)$$ over $G/P=\mathbb{P}^1$. This is a $G$-equivariant line bundle, and the $G$-action alone tells you that $Aut(G/P,L)$ has dimension greater than $1$. So, it seems that $Aut(G/P,L)/S^1$ cannot possibly be discrete.
Furthermore, the bundles $\mathcal{O}(n)$, $n\in\mathbb{Z}$, are (up to isomorphism) all of the holomorphic line bundles on $G/P=\mathbb{P}^1$. So, it seems your condition cannot be satisfied for the $G$ and $P$ I have chosen. For this reason, I am skeptical as to whether it holds for higher-dimensional partial flag varieties.
ADDED: I think I have a general idea for a proof. Let $G$ be connected, simply-connected and semisimple. First note that every holomorphic line bundle $L$ over $G/P$ is actually isomorphic to an associated line bundle $G\times_{P}\mathbb{C}(\alpha)$, where $\alpha$ is a root orthogonal to every simple root defining $P$. However, $L=G\times_P\mathbb{C}(\alpha)$ is a $G$-equivariant line bundle. This $G$-action is a lift of the $G$-action on the base $G/P$, so the $G$-action alone tells you $Aut(G/P,L)$ has dimension greater than $1$. As above, this seems to imply $Aut(G/P,L)/S^1$ cannot be discrete.
