This may be an elementary question, but I'm having trouble coming up with an answer: Let $\tilde{N} = T^*(G/B)$ be the Springer resolution of the nilpotent cone. Does it have finitely many $G$-orbits? If not, what's the "first" example where it doesn't?

The answer for $SL_2$ is obviously yes (finitely many orbits). The answer for $SL_3$ also seems to be yes. There are three nilpotent orbits. The Springer fiber above the zero nilpotent orbit is $G/B$ and has one orbit associated to it. The fiber above the regular orbit is a point, so it has one orbit associated to it. Above a subregular element, say $x = \left(\begin{matrix} 0&1&0\\0&0&0\\0&0&0\end{matrix}\right)$, the fiber is given by flags (the notation here is: take the span of the first column, then the span of the first two columns etc.) $$\left(\begin{matrix} a&1&0\\0&0&1\\b&0&0\end{matrix}\right), \left(\begin{matrix} 1&0&0\\0&a&1\\0&b&0\end{matrix}\right)$$ i.e. two $\mathbb{P}^1$ intersecting nodally. However, the torus $$\left(\begin{matrix}t&0&0\\0&t&0\\0&0&t^{-2}\end{matrix}\right)$$ stabilizes $x$, and acts on this fiber by five orbits. This seems to indicate to me that there are only finitely many orbits for $SL_3$. Same seems to be true for $SL_4$, though I'm less certain about my calculation there.

On the other hand, I can't think of a reason why this would be the case, and can't seem to find anything about it in the literature (which leads me to believe it's possibly untrue, and I'm missing something obvious).