For a continuous action, the answer is yes: $\mathbb{C}-\\{0\\}=\mathrm{GL}(1,\mathbb{C})$ is a classical Lie group which acts transitively on itself by multiplication. Now use that $\\{z\in\mathbb{C}:r<|z|
For a holomorphic action, the answer is no: If $f$ is a biholomorphism of the annuli, then either $|f(z)|=|z|$ for all $z$ or $|f(z)|=Rr/|z|$ for all $z$ (assuming $0