Is the solution to the functional equation $\widehat{F}(z) = z\widehat{G}(\widehat{F}(z))$ unique? I am reading Martin Aigner's A Course in Enumeration, and in $\S$3.3 The Exponential Formula, the author states and proves the following theorem (on page 117):

Theorem 3.8. Suppose $F(z) = zG(F(z))$, $G(0) \neq 0$. Then
$$
[z^n] F(z) = \frac{1}{n} [z^{n-1}]G(z)^n.
$$

Here, $F$ and $G$ are formal power series over $\Bbb{C}$ in the variable $z$. As a corollary, the author establishes the Lagrange Inversion Formula.
The proof of Theorem 3.8 starts as follows:

Proof. We write $F(z)$ and $G(z)$ in exponential form, $\widehat{F}(z) = \sum_{n \geq 1} f(n) \frac{z^n}{n!}$, $\widehat{G}(z) = \sum_{n \geq 0} g(n) \frac{z^n}{n!}$. For a rooted tree $T$ on $\{1,\dotsc,n\}$ let
  $$
g^T := g(0)^{r_0} g(1)^{r_1} g(2)^{r_2} \dotsm,
$$
  where $r_i$ is the number of vertices in $T$ with out-degree $i$ (edges pointing away from the root). The sequence $(r_0,r_1,r_2,\dotsc)$ is called the type of $T$. Since $T$ has $n-1$ edges, we have
  $$
\sum_{i \geq 0} r_i = n, \quad \sum_{i \geq 0} ir_i = n-1.
$$
  Let $f(n) = \sum_T g^T$ over all rooted trees on $\{1,\dotsc,n\}$.

Then, the following claim is stated:

Claim 1. $\widehat{F}(z) = \sum_{n \geq 1} f(n) \frac{z^n}{n!}$ is the solution of the functional equation $\widehat{F}(z) = z\widehat{G}(\widehat{F}(z))$.

In the proof of the claim, the author shows that if we take $\widehat{F}(z) = \sum_{n \geq 1} f(n) \frac{z^n}{n!}$, where $f(n) = \sum_T g^T$, then $\widehat{F}(z)$ does satisfy the functional equation $\widehat{F}(z) = z\widehat{G}(\widehat{F}(z))$. However, no comment is made regarding uniqueness.
Is it obvious that this particular choice of $\widehat{F}(z)$ is indeed the solution to the functional equation $\widehat{F}(z) = z\widehat{G}(\widehat{F}(z))$? How can I see this?

The rest of the proof is easy to follow. After establishing Claim 1, we have that $[z^n]F(z) = \frac{f(n)}{n!}$, so it only remains to show that
$$
f(n) = (n-1)! [z^{n-1}]G(z)^n.
$$
For this the following claim is stated and proved.

Claim 2. There are precisely $\binom{n-1}{d_1 d_2 \dotso d_n}$ rooted trees on $\{1,\dotsc,n\}$ in which vertex $i$ has out-degree $d_i$, $\sum_{i=1}^n d_i = n-1$.

The proof of Theorem 3.8 is easily completed from here.
 A: I understood why the solution must be unique a few minutes after posting the question. Below is my answer to the question; other viewpoints are also welcome.

Since $G(0) \neq 0$, the power series $G(z)$ is invertible in the ring $\Bbb{C}[[z]]$, say with inverse $K(z)$ (that is, $1/G(z) = K(z))$. Note that as power series, $G(z) = \widehat{G}(z)$, so we also have $1/\widehat{G}(z) = K(z)$.
Now, suppose that $\widehat{F}(z)$ is a solution to the functional equation $\widehat{F}(z) = z\widehat{G}(\widehat{F}(z))$. Then, multiplying both sides by $K(\widehat{F}(z))$, we get
$$
\widehat{F}(z) K(\widehat{F}(z)) = z.
$$
Hence, if $P(z)$ is the power series $zK(z)$, then $P$ is the compositional inverse of $\widehat{F}$, that is, $P(\widehat{F}(z)) = z = \widehat{F}(P(z))$. This is because (exercise):


*

*any left compositional inverse is also a right compositional inverse (and vice-versa), and

*if a compositional inverse exists, then it is unique.


Hence, if $\widehat{F}(z)$ is a solution to the functional equation $\widehat{F}(z) = z\widehat{G}(\widehat{F}(z))$, then it is uniquely determined as the compositional inverse of the power series $zK(z) = z/G(z)$.
