Let $\mathfrak{I}:=\big\{ \, f:=\sum\_{k=0}^\infty f\_k z^k \in\mathbb{C}[[z]]\; : \text{s.t. }\; f\_0=0 \;\text{ and }\; f\_1=1\big\}$. A most basic result about linearization states that, for any $f\in\mathfrak{I}$ and for any $\lambda\in\mathbb{C}$ not a root of unity, there exists a unique $h\in\mathfrak{I}$ that linearizes $f(\lambda z)$:
$$h(\lambda z)=f(\lambda h(z)).$$
Existence and uniqueness of this conjugation in the setting of formal power series is indeed easily established, as the coefficients of $h$ can be determined inductively from the data $(f\_1,f\_2,\dots) $ expanding the composition: one finds
$$h\_1=1$$
$$h\_{n+1}={1\over \lambda^{n}-1 } \sum\_{k\ge2,\,j} f\_k \lambda^{k-1} h\_{j\_1}\dots h\_{j\_k},$$
the latter sum being extended over all integers $k\ge2$ and multi-indices $j\in\mathbb{Z}\_+^k$ of length $|j|:=j\_1+\dots+j\_k=n+1$, thus with all components $j\_i\le n$.
The above recursion implies that the coefficients $h\_n$ should be rational functions of $\lambda$, with denominators in the form $\prod\_{k=1}^{n-1}(\lambda^k-1)^{m\_{n,k}}$, where the exponents $m\_{k,n}$ may be larger than $1$, at least at a first glance. However, after experiments with various $f$ and up to $n=50$ I observed that the form of $h\_n$ always shows the simpler denominator, with all $m\_{k,n}=1$:
$$h\_n={P\_n(\lambda)\over (\lambda-1)(\lambda^2-1)\dots(\lambda^{n-1}-1)},$$
for some polynomials $P\_n(\lambda)$ of degree at most ${n\choose 2}$.



> 
> Is this really a general phenomenon? If so, **why**? Is there a simpler formula (maybe still recursive) for $h\_n$ than the above recursion?
> 
> 
> 



Linearizing a power series by conjugation
You may find useful information in a recent article by D. Sauzin and al. "Explicit linearization of one-dimensional germs through tree-expansions" [here](https://hal.archives-ouvertes.fr/hal-01053805/), where they use "mould calculus" (introduced by J. Écalle 40 years ago) to write down the coefficients $h\_n$ and explore the combinaotrial structure of their family. 


