Would a Schwarzschild black hole "appear" to be a sphere in all reference frames? I am wondering if all observers would agree that a non-rotating, non-charged black hole is spherical (i.e. there is no reference frame where one would measure it to be oblong in one dimension). I assume that this is true, but with relativity I have learned not to assume.
 A: Even in special relativity, the optical appearance of an ordinary sphere such as a basketball is that it is always a circle (not an ellipse) in all frames of reference and from all points of view, although areas within the surface of the sphere are distorted in size. The special-relativistic length contraction is not the same as what you actually see in optical observations. This video has some nice simulations near the end: http://youtube.com/watch?v=JQnHTKZBTI4
So now let's talk about a black hole. General relativity does not have the concept of a global frame of reference, so there is no way to say what the shape of the event horizon of a black hole is in some frame of reference. Therefore all we can really talk about is the optically determined silhouette; the notion of a Lorentz contraction can't be applied here. I believe it's true that in optical observations, the silhouette of a Schwarzschild black hole's event horizon is circular for all observers in optical observations. See What will the universe look like for anyone falling into a black hole? for some simulated views with radial motion. Of course the real question that arises is not for radial motion but for tangential motion. I think the lecture by Riazuelo (in French) linked to from my answer does discuss this, although I haven't made an intensive attempt to figure out all the French. (My French is pretty weak.)
A: As long as the observer is stationary, then yes. The Schwarzschild metric has complete spherical symmetry: in fancy terms, you can say that given any two timelike trajectories at fixed values of $(r, \theta, \varphi)$ and at the same $r$, there is always an isometry taking one to the other. You can also calculate the shape of the black hole's shadow: it turns out to be a circle, a bit larger than the black hole itself, and independent of the observer's position.
However, you are right to be wary of relativity, because for a Kerr (i.e., rotating) black hole this is not true! The event horizon itself is kind of ellipsoid shaped, so obviously it doesn't look the same from all angles, but there are optical effects that make it look squished, but only when seen from the side. When seem from the top it is circular, like you would expect for an ellipsoid, but as you look from different angles you'll find that one side gets flatter, the other gets longer, and the whole thing moves to the side. This happens because the effect the black hole's gravity has on light rays depends on which direction they're rotating in.
