Let $G$ be a compact connected Lie group and $\mu:T\to S^1$ be a representation of a maximal torus $T \subset G$ and $\lambda=d\mu$ be a weight for some $\lambda\in\mathfrak{t}^*$ (where $\mathfrak{t}$ is the Lie algebra of the maximal torus). Let $O_{\lambda}$ be a generic coadjoint orbit and $O_{\lambda'}$ be another coadjoint orbit ($\lambda'\in \mathfrak{g}^*$). When are the two representations $\rho \colon G \to GL(\Gamma(O_{\lambda},G\times_{\mu} \mathbb{C}))$ (where $\Gamma(O_{\lambda},G\times_{\mu} \mathbb{C})$ is the set of holomorphic sections) and $\rho \colon G \to GL(\Gamma(O_{\lambda'},G\times_{\xi} \mathbb{C}))$ (where $\xi:O_{\lambda'}\to S^1$ is a character) equivalent?
These representations are the same if the coadjoint orbits are the same. Your notation is a bit confusing, but the basic fact is this: every coadjoint orbit intersects the positive Weyl chamber in a single point. If that point is an integral weight, then by BorelWeil, the holomorphic sections of the line bundle you've written down are the irreducible representation with that highest weight. If the point isn't integral, then there simply is no such line bundle (since the weight don't integrate to the torus), which corresponds to the fact that no finite dimension representation has that weight.
I think the representations are equivalent if and only if $\lambda$ and $\lambda'$ are conjugate under the Weyl group. To see this, note that the irreducible representations of $G$ can be parametrized by dominant weights, or equivalently, Weylorbits of weights.

$\begingroup$ Thanks Peter, I think you speak about BorelWeil theorem, can you give a referrence for the first part of your answer? $\endgroup$ – user21574 Feb 3 '14 at 18:48

1$\begingroup$ Hi, Hassan. Yes, I am using the BorelWeil(Bott?) Theorem. The first part of my answer implicitly refers to the Theorem of the Highest Weight for representations of complex reductive/compact groups. $\endgroup$ – Peter Crooks Feb 3 '14 at 18:53

1$\begingroup$ It's not clear what it would mean for $\lambda$ and $\lambda'$ to be conjugate under the Weyl group, since $\lambda'$ isn't assumed to be $\mathfrak{t}^*$ (I think; the notation in the question is a mess, so it's hard to be sure). If $\lambda'$ is in $\mathfrak{t}^*$ and conjugate to $\lambda$ then they live in the same coadjoint orbit. $\endgroup$ – Ben Webster♦ Feb 5 '14 at 19:08

$\begingroup$ I admit that I hastily assumed $\lambda'\in\mathfrak{t}^*$. Of course, one can write any coadjoint orbit as the coadjoint orbit containing some $\lambda'\in\mathfrak{t}^*$. Since $\mathcal{O}_{\lambda}$ and $\mathcal{O}_{\lambda'}$ coincide if and only if $\lambda$ and $\lambda'$ are Weylconjugate, we have arrived at exactly the same answer. $\endgroup$ – Peter Crooks Feb 5 '14 at 20:50