Vtyjkyuk

Here is a really nice video that shows the following:

Suppose we have a subset of $mathbbCcup infty$. Then we can project it onto the 2-sphere via the inverse stereographic projection. If we rotate or translate the sphere in the surrounding $mathbbR^3$ and apply the stereographic projection again, then the result amounts to a Möbius transformation of the initial subset. Here is a screenshot to illustrate this:

$hskip2in$

Furthermore, on this wiki-page on conformal maps, the 4th paragraph of the subsection "Complex analysis" says that "A map of the extended complex plane (which is conformally equivalent to a sphere) onto itself is conformal if and only if it is a Möbius transformation."

Now this seems quite interesting to me because it seems to imply that if I take a subset of $mathbbR^2$ and apply a conformal transformation $Phi$ to it, then I can always write it as
beginequation
Phi = Pcirc phicirc P^-1
endequation
where $P$ is the stereographic projection and $phi$ is a rotation and/or translation of the sphere in 3-space.

My first question is: Is that right?

My second question concerns whether this generalises to higher dimensions. I saw the wiki page on Liouville's theorem which states that every conformal mapping in $mathbbR^n$, $nge 3$ is a Möbius transformation.
However, I am not sure if a Möbius transformation in higher dimensions still has the same intuitive meaning as in 2 dimensions.

Here is a counterexample. Let $Q$ be the open first quadrant in $mathbbC$, let $H$ be the open upper half plane of $mathbbC$, let $f : Q to H$ be the conformal transformation defined by $f(z)=z^2$. I'll abuse notation and write the inverse transformation $f^-1 : H to Q$ as $f^-1(z) = sqrtz$. Consider the following transformation:
$$Phi : Q xrightarrowz mapsto z^2 H xrightarrowz to z+1 H xrightarrowz mapsto sqrtz Q
$$
and so we may write $Phi(z) = sqrtz^2+1$. This map $Phi$ does not extend to a Möbius transformation of the extended complex plane. So, if you conjugate by stereographic projection to get $phi = P^-1 circ Phi circ P$, then $phi$ does not extend to any of the particularly nice functions on the 2-sphere that you ask about.

In general, conformal transformations are much wilder beasts than Möbius transformations. You might be interested in the Schwartzian Derivative, a quantity which, given a conformal transformation between two open subsets of the complex plane, let's you measure whether that transformation is "infinitesmally Möbius". Calculating the Schwartzian derivative of $sqrtz^2+1$ will show that it is not even infinitesmally Möbius, let alone the restriction of a Möbius transformation.

What you seem to have overlooked is what was mentioned in the comment of @Rahul: that theorem quoted from the wiki-page only applies to conformal bijections of $mathbbC cup infty$.

(This answer was written before the large edit of the question.)

In defining a Moebius transformation $Tincal M$, $ T(z)=az+bover cz+d$, you have three complex, or six real, degrees of freedom, whereas in defining a rotation of $S^2$ you have just three real degrees of freedom: You can choose the axis, and then the rotation angle. This argument already shows that there are "many more" Moebius transformations than there are rotations of $S^2$. (It is unclear to me what you mean by a translation of $S^2$.)

Where are the three real degrees of freedom lost? Number one, a Moebius transformation may have any two points $a$, $binbar mathbb C$ as fixed points (there are even $Tincal M$ with just one fixed point), whereas the two fixed points of a rotation stereographically correspond to complex $a$, $b$ satisfying $abar b=-1$. Number two, for a general $Tincal M$ with two fixed points one is attracting, and the other repelling, whereas for a rotation the corresponding eigenvalue satisfies $|lambda|=1$.