Would a stationary observer see a circle, moving at high speeds near $c$, as an ellipse? "Any Lorentz transformation corresponds to a bilinear transformation of this ratio and therefore to a projective (conformal)transformation of the sphere, which sends circles into circles."
How is this? Is there an intuitive explanation?
http://reprints.gravitywaves.com/GR/Penrose-1960_ASpinorApproachToGeneralRelativity_AnnsPhys.pdf
 A: There are different ways of using the term "see" and "observe" in relativity theory and indeed in physics generally: what they all mean is acquire information by some method, and the relevant point here is that depending on the method employed, what that observer will "see" will vary.
The Lorentz-FitzGerald contraction is only observed when one is using a "seeing" method that can "see" a Lorentz present instant. This is actually not something that is possible to achieve with a local observer - it would either require tachyons (which do not appear to exist in real life), or else it would necessitate an observing apparatus that is both spread out over a large finite area of space in which observations are to be conducted and, moreover, which can accumulate data over time, so that the Lorentz observer actually does not register things as happening simultaneously with their actual happenings, but instead must reconstruct those events from the collected data at a point in the future. Using such a system, yes, the speeding circular object appears elliptical, and a speeding spherical object (in 4D spacetime) appears ellipsoidal.
What Penrose looks to be talking about, though, given he refers shortly before your quoted point to a "field of view", is literal seeing - i.e. what you would see using your actual physical eyes or, at least, a suitably high-speed camera, that is sensing scattered or emitted electromagnetic radiation. This is a considerably different method, but entirely possible to model mathematically. A camera or eye sees a (part of a) projection onto a sphere - the view sphere - of the intersection of its past light cone with the spacetime worms of objects in distant spacetime.
In particular, you can imagine rays (not light, but "lines of sight") leading out from the viewer's position at the apex of the cone and along its surface. If such a ray hits something, then the viewer sees the hit point at that point on their view circle and, of course, not what is behind it. In a 4D spacetime so the light cone is a hypercone, you get the same thing but with a view sphere.
And what Penrose shows is that the speeding spherical object does not appear, to this observer, as contracted, but rotated. This effect is sometimes called the "Penrose-Terrell rotation" (after James Terrell, who also calculated it independently of Penrose). For non-spherical objects, the effect looks like a more complex distortion - it only looks as a rotation for spheres. There are plenty of simulation animations on Youtube showing the effect with a complex environment - overall, it strongly resembles looking through a fisheye lens, but can "see around walls" in a way that no such lens could do by itself, and when that "seeing around walls" happens to a sphere, the maths comes out "just so" that it has the same apparent effect as a rotation of said sphere around an axis parallel(!) to the observer's vision.
Which one is the "right" observation? The answer is both and neither: they both define coordinate systems on spacetime, indicating and labelling events in a certain way based on how they appear configured to each kind of observer, and those coordinate systems admit a valid description of physics in that all statements about cause and consequence will be identical when suitably translated; it is just that in the Lorentz frames, it is mathematically simplest. The Lorentz frames are not preferred by spacetime (and thus not per se a feature of "relativity" by itself), but by dynamics - dynamics is such that non-interacting objects will be either at rest or else following straight line paths in such frames, no matter how they are configured. And it also is much easier to talk geometrically of a simple contraction (squash transformation) in one direction than a complex fisheye modification.
A: The Sympathizer is right; I shall offer a simple explanation.
First let's consider a visual effect which has no particular relation to Special Relativity, it is also found in Newtonian physics. It is a visual elongation of the appearance of a moving object. Here is why. Suppose a rectangular object has sides of lengths $L_x$, $L_y$ and is sitting still at a distance $y$ from you. Then it subtends an angle of about $L_x/y$
(or more precisely $2 \tan^{-1}(L_x/2y)$). You, looking at it from that distance, see it filling up this angle in your field of view.
Now suppose the object moves in the $x$ direction, along the line $y=$ constant, while you stand at the origin. In other words it passes from left to right in front of you. Think about the waves emitted from all points on this rectangular object. The side of the rectangle nearer to you is at distance $y$ when it just reaches the point in front of you, and the opposite side is further away, at distance $y+L_y$. So light waves arriving at you from the nearer side at $t=0$ must have been emitted at the time $t_1 = -y/c$, and light waves arriving at you from the far side must have been emitted at time $t_2 \simeq -(y+L_y)/c$. Between these two times the rectangle moves through a distance $x = v(t_1 - t_2) = v L_y/c$. Therefore the light waves arriving at the origin at $t=0$ came from a set of locations which are spread along the $x$ direction by about
$$
L_x + (v/c) L_y.
$$
The angle subtended at the origin by this set of locations is about
$$
\frac{L_x + (v/c) L_y}{y}.
$$
This is larger than $L_x/y$ so the object appears elongated. This is just a question
of motion and angles and is an entirely Newtonian (not relativistic) effect. If you like you can avoid light and treat the scenario using sound waves or water waves and get the same answer (where $c$ is the speed of the waves).
The formula above is a little rough because I did not want to get into details of inverse tangent of angles when the object is not directly in front of you.
Now let's bring in relativity. According to relativity, when it is in motion the object is contracted along its direction of motion by the factor
$$
\gamma = \frac{1}{\sqrt{1-v^2/c^2}} \simeq 1 + \frac{v^2}{2c^2}
$$
where the approximate version is for $v \ll c$. This is 'Lorentz contraction'. What does it mean exactly? It refers to the spatial location of all points on the object at some moment in time in the reference frame relative to which it moves at speed $v$.
The combination of the relativistic contraction and the visual appearance of elongation combine to make the angle subtended by the moving object be about
$$
\frac{L_x/\gamma + (v/c) L_y}{y} \simeq \frac{L_x + (v/c)L_y - (v^2/2c^2)L_x}{y}.
$$
For a square object the two effects compensate out and we find that the object still looks elongated but by less than the Newtonian calculation predicts. To be precise, it does not look elongated so much as rotated, and the diagonal of a rectangle is longer than the sides.
The Newtonian effect means it should look longer, but the Lorentz contraction says it is shorter, and the two effects combine.
What Penrose realized is that these effects balance out exactly in the case of a circle (or a sphere). Then it is exact after you allow for all sine and tangent functions and the exact value of $\gamma$. Notice that the effect involves both the width and depth of the object in view. It would not occur for a pencil-shaped rod parallel to the $x$ axis and moving in the $x$ direction. Such a rod would be Lorentz contracted and its visual appearance (the angle it subtends at the origin while it whizzes by) would be smaller than $L_x/y$ by a factor $\gamma$.
Also, in the case of a spherical object, any pictures painted on the surface of the sphere would look distorted to you the viewer as the sphere passes in front of you.
