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?