Koenigs Linearization Theorem: If the magnitude (absolute value) of the multiplier $\lambda = \dot{f} (0)$ of a holomorphic map $f$ is not strictly equal to 0 or 1, that is $λ \ne \{0, 1 \}$, then a local holomorphic change of coordinates $w = \phi(z)$, called the Koenig's function, unique up to a scalar multiplication by nonzero constant, exists, having a fixed-point at the origin $\phi (0) = 0$ such that Schrοder's equation is true $\phi \circ f \circ \phi^{- 1} = \lambda w \forall w \in \varepsilon_0$ for some neighborhood $\varepsilon_0$ of the origin 0.
The Simplicity Lemma: The Koenig's function $\phi$ is the only solution to the eigenvalue equation as all of the other solutions are constant multiples of powers of the principal eigenfunction $\sigma(t)$ of $C_\phi$
My question is: if the (Schroeder iteration) function φ is the Newton map $$t-\frac{f(t)}{\frac{d}{dt}f(t)}$$ does the simplicity lemma tell us anything about the multiplicity of the root of f which is the fixed-point to which the Koenigs function converges?
It does not tell us that the root of f has multiplicity one because for instance if we let $f (t) = \tanh (\ln (1 - t^2))$ then its easy to see that f has a double-root at the origin (factor the function into a rational function of two finite products $$f(t)=\frac{(1-t^2)^2-1}{(1-t^2)^2+1} $$
However, if we take the composite function $g(t)=f(g(t))$ then the mapping $f(t)$ converts roots of multiplicity 1 to roots of multiplicity 2, and thus makes Koenig linearization applicable to functions with super-attractive fixed-points at the points where there are roots of multiplicity one .
Does there exist some way to say that if the Newton map of the composite function $f(g(t))$ has a geometrically attracting fixed-point of multiplier 1/2 then the root at that point of $g(t)$ indeed is simple with multiplicity one?