I know that for real manifolds, being diffeomorphic does not imply having same differential structure. For example $ \mathbb R $ with atlas $\{ (\mathbb R , x) \}$ and $ \mathbb R $ with atlas $\{ (\mathbb R , x^3) \}$ are diffeomorphic but two charts are not compatible.
Is it similar for biholomorphic riemann surfaces?
i.e. does there exist biholomorphic Riemann surfaces having holomorphically uncompatible charts?
The same basic idea as $x^3$ vs $x$ can produce a huge number of examples - we can modify the holomorphic structure by composing it with any nasty homeomorphism. In detail:
Let $(\Sigma,A = \{\phi_i\})$ be any Riemann surface; i.e. a topological surface $\Sigma$ equipped with an atlas of homeomorphisms $\phi_i : \Sigma \supset U_i \to \phi_i(U_i) \subset \mathbb C$ such that $\phi_i \circ \phi_j^{-1}$ is holomorphic wherever defined. Let $\psi : \Sigma \to \Sigma$ be any homeomorphism that is not holomorphic as a map $(\Sigma,A) \to (\Sigma,A).$
I claim that we can replace each $\phi_i$ with $\tilde\phi_i = \phi_i \circ \psi$ to obtain a holomorphic atlas $\tilde A$ on $\Sigma$ which is biholomorphic to the original one, but not compatible.
First, on the overlaps we have $$\tilde \phi_i \circ \tilde \phi_j^{-1} = \phi_i \circ \psi \circ \psi^{-1} \circ \phi_j^{-1} = \phi_i \circ \phi_j^{-1},$$ so the fact that $A$ is an atlas tells us that $\tilde A$ is an atlas.
Secondly, $\psi$ is a biholomorphism $(\Sigma, \tilde A) \to (\Sigma, A):$ for any charts $\tilde \phi_i, \phi_j$ we have $$ \phi_j \circ \psi \circ \tilde \phi_i^{-1} = \phi_j \circ \psi \circ \psi^{-1} \circ \phi_i^{-1} = \phi_j \circ \phi_i^{-1},$$ which again is holomorphic because $A$ is a holomorphic atlas.
Finally, the fact that $\psi : (\Sigma,A) \to (\Sigma,A)$ is not holomorphic tells us that there are some overlapping charts on which $\phi_i \circ \psi \circ \phi_j^{-1}=\tilde \phi_i \circ \phi_j^{-1}$ is not holomorphic, so $\tilde \phi_i$ is not compatible with $A$.