Fixed point of holomorphic map has a constant multiplier

Let $M$ be a Riemann surface, and let $f : M \to M$ be a holomorphic map. If $f$ has a fixed point $z_0 \in M$, we define the multiplier of $z_0$ to be the quantity $$ \lambda = \frac{d(\phi \circ f \circ \phi^{-1})}{dz}(\phi(z_0)), $$ where $\phi$ is a coordinate chart of $M$. This answer states that the multiplier $\lambda$ is well-defined, in the sense that $\lambda$ is independent of the choice of $\phi$. How can I prove this?

Here's an example that helps indicate that the above statement might be true. Let $\overline{\mathbb{C}}$ denote the Riemann sphere, which we equip with the coordinate charts $$ \phi_1 : \mathbb{C} \to \mathbb{C}, \quad \phi_1(z) = z, \\ \phi_2 : \overline{\mathbb{C}} \setminus \{0\} \to \mathbb{C}, \quad \phi_2(z) = \frac{1}{z}. $$ Let $f : \overline{\mathbb{C}} \to \overline{\mathbb{C}}$ be the squaring map $f(z) = z^2$. Then $1$ is a fixed point of $f$, and $$ \frac{d(\phi_1 \circ f \circ \phi_1^{-1})}{dz}(\phi_1(1)) = \frac{df}{dz}(1) = \frac{d(\phi_2 \circ f \circ \phi_2^{-1})}{dz}(\phi_2(1)), $$ which proves the multiplier of $1$ is precisely $f'(1) = 2$.

If $\phi$ and $\psi$ are two charts in a neighborhood $U(z_0)$ then $\psi = h\circ \phi$ with a holomorphic function $h$. It follows from the chain rule that $$ \frac{d(\psi \circ f \circ \psi^{-1})}{dz}(\psi(z_0)) = \frac{d(h \circ \phi \circ f \circ \phi^{-1} \circ h^{-1})}{dz}(h(\phi(z_0))) \\ = h'(\phi(z)) \cdot \frac{d(\phi \circ f \circ \phi^{-1})}{dz}(\phi(z_0)) \cdot (h^{-1})'(h(\phi(z_0)) \, . $$ Using the rule for the derivative of an inverse function we see that the product of the first and the last factor on the right is equal to one, and it follows that $$ \frac{d(\psi \circ f \circ \psi^{-1})}{dz}(\psi(z_0)) = \frac{d(\phi \circ f \circ \phi^{-1})}{dz}(\phi(z_0)) \, . $$