Most general linear transformation of $|z|=r$ into itself using cross ratio This question (without the cross ratio part) was asked earlier today, as well as a few times before.  Here was the question that was asked earlier today: Find all Möbius transformations that map the circle $|z|=R$ into itself
Now, I am wondering if this problem could instead be set up using the cross ratio, but I suppose I am having a hard time setting everything up.
So, I want $(z,z_1,z_2,z_3)=(w,w_1,w_2,w_3)$, where $w=f(z)$.
Now, my thoughts, and this is where I suppose I am going wrong, I suppose that we would have $(z,0,1,a)=(w,0,1,\infty)$, which, after setting up the cross ratio and solving for $w$ would give $w=\frac{z(a-1)}{-z+a}$.  But, I believe this is wrong.  For instance, I have that the center of the original circle (the origin) is mapped backed to the origin.  But, I don't believe this to be necessary since we want the MOST GENERAL linear transformation.  Also, I am mapping $1$ to $1$ based on some of the other answers that were given.
My question is in two parts: first, is this a problem we could do using the cross ratio, and second, if the first part is doable, can you provide some insight into how you are going about setting up the "$(z,z_1,z_2,z_3)=(w,w_1,w_2,w_3)$" bit?
Thank you!!
 A: Consider the case $r=1$ first, i.e. $f$ is a Möbius transformation which maps the unit circle $|z|=1$ onto itself.
Let $f^{-1}(0) = a$. Then, because $f$ preserves symmetry with respect to the unit circle, $f^{-1}(\infty) = 1/\overline a$. The image of a third point determines $f$ uniquely, so let us set $c = f(1)$. Note that $|c| = 1$. $f$ preserves the cross ratio, so we can conclude that
$$ \tag{*}
 (z, 1, a, 1/\overline a) = (f(z), c, 0, \infty)
$$
and we get
$$
cf(z) = \frac{z-a}{z-1/\overline a} \cdot \frac{1-1/\overline a}{1-a} \\
\iff f(z) = \frac 1c \frac{\overline a - 1}{a-1} \cdot \frac{z-a}{1-\overline a z} \, .
$$
The factor $\frac 1c \frac{\overline a - 1}{a-1}$ has absolute value one, therefore
$$ \tag{**}
 f(z) = e^{i \lambda} \frac{z-a}{1-\overline a z}
$$
for some $\lambda \in \Bbb R$ and some $a \in \Bbb C$ with $|a| \ne 1$.
So any Möbius transformation which maps the unit circle onto itself is necessary of the form $(**)$.
On the other hand, if $f$ is defined by $(**)$ then $f$ satisfies $(*)$ with some $c$ of absolute value one, which implies that $f$ maps the unit circle onto itself.
(Depending on whether $|a| < 1$ or $|a| > 1$, $f$ maps the interior of the unit circle to the interior or to the exterior of the unit circle. The case $|a| < 1$ gives exactly the conformal automorphisms of the unit disk.)

For arbitrary $r > 0$ you can consider the mapping $\tilde f(z) = f(rz)/r$ which must be of the form $(**)$, or repeat the above argument with mirroring at the circle $|z|=r$:
$$
 (z, r, a, r^2/\overline a) = (f(z), c, 0, \infty) \, .
$$
