Let $C$ be the circle $\mid z \mid =r$ and $D$ be the circle $\mid z \mid =R$ with $0<r<R$.
Then when does there exist a conformal map that takes $D$ and its interior to the closed ball $\mid z \mid \leq 2$ and simultaneously take $C$ and its interior to the closed ball $\mid z \mid \leq 1$?
Upon looking at the automorphisms of the unit disc, I conclude that there exists such a conformal map if and only if $R/r =2$. Am I correct? Could anyone please check me and provide me with a feedback?
You are correct, you need the ratio $R/r=2$. In fact, a conformal map preserves the `modulus' of the annulus $A= \{z \in \mathbb{C}: \; r< |z| <R\}$, which is defined as $\operatorname{Mod} (A) = \frac{1}{2\pi}\log \frac{R}{r}$. This is result from complex analysis that is proved in several places, for example here, or on Wiki.
If you are already familiar with complex analysis, here is a sketch of one way to proceed: Suppose $f:A \to A'$ is a conformal map between annuli (centered at the origin). By repeated Schwarz reflection, you can extend to a conformal map $f:\mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\}$. Show that the singularity at $0$ is removable. Then you know that the extended map is a conformal automorphism of the plane and hence of the form $f(z) = az+b$. Since it fixes the origin, we can conclude $b=0$ and the ratios of the inner and outer radii of $A$ and $A'$ must be equal.