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 $|z|=r$:
$$
(z, r, a, r^2/\overline a) = (f(z), c, 0, \infty) \, .
$$
Best Answer
The linear transformation you are looking for is of the form: $$ \varphi(z)=\frac{\omega z}{2}-1, $$ where $\lvert\omega\rvert=1$.
Verification. If $\lvert z\rvert=2$, then $z=2\mathrm{e}^{it}$, and $\varphi(z)=\omega\mathrm{e}^{it}-1$, which is the parametrical description of the circle $\lvert z+1\rvert=1$.
Determination of $\omega$. It is going to be determined by the restriction $$ \varphi(-2)=0, $$ equivalently $$ 0=\frac{\omega(-2) }{2}-1=-(1+\omega), $$ and hence $$ \omega=-1. $$ Thus the linear transformation is $$ \varphi(z)=-\frac{z}{2}-1, $$ Clearly $\varphi(0)=-1$.
Apparently, such linear transformation does not exist.