function $f$ is conjugated by a Möbius transformation, fixed points When a function $f$ is conjugated by a Möbius transformation $h$, i.e. $g \circ h = h \circ f$ any fixed point  of $f$ is mapped to a fixed point of $g$ and the multiplier is equal.
I understand why the fixed point is mapped to a fixed point but I don't understand why the multipliers are equal from this proof. Can someone explain?
If $z_0$ a fixed point of $f$ and $w_0 = h(z_0)$ a fixed point of $g$ then $$f(z_0) = h^{-1}(g(h(z_0))) = z_0$$ thus $g(h(z_0)) = h(z_0)$, that is, $g(w_0)=w_0$ and $w_0$ is a fixed point of $g$ as required. Now, by the chain rule $$g'(w_0 ) = h'(w_0 ) f'(z_0 ) (h^{-1})'(w_0 )$$ and by the formula for derivative of an inverse function obtain $ g'(w_0)=f'(z_0) $
The formula for the inverse derivative gives $$g'(w_0) = h'(w_0)f'(z_0)\frac{1}{h'(z_0)}$$ but how is this true? How do we show $h'(w_0)=h'(z_0)$ so that they cancel and we obtain the equality?

If you take the derivatives of $h(f(z))=g(h(z))$ by the chain rule then
$$
h'(f(z))f'(z)=g'(h(z))h'(z)
$$
and inserting the fixed points gives
$$
h'(z_0)f'(z_0)=g'(w_0)h'(z_0)
$$
which up to this point is true for vector valued functions. For complex functions the factors $h'(z_0)$ on each side cancel. In the vector case the matrices $f'(z_0)$ and $g'(w_0)$ are conjugate so that their Jacobi normal forms are identical.
