Post History

If we look at formal power series and ignore questions of convergence for now, we can take $f(x) = \sum_{i \ge 0} a_i x^i$. Then the question is which sequences of $a_i$ satisfy $$\sum_{j \ge 0} a_j \left(\sum_{i \ge 0} a_i x^i\right)^j = x$$

The case $a_0 \neq 0$ is awkward, because we immediately get the constraint $$f(f(0)) = \sum_{i \ge 0} a_i a_0^i = 0$$ which involves all of the coefficients. But if we focus on the case $a_0 = 0$ we get a series of finite constraints:

$$\begin{eqnarray*}
a_1^2 &=& 1 \\
a_1^2 a_2 + a_1 a_2 &=& 0 \\
a_1^3 a_3 + 2 a_1 a_2^2 + a_1 a_3 &=& 0 \\
a_1^4 a_4 + 3 a_1^2 a_2 a_3 + a_2^3 + 2 a_1 a_2 a_3 + a_1 a_4 &=& 0 \\
a_1^5 a_5 + 4 a_1^3 a_2 a_4 + 3 a_1 a_2^2 a_3 + 3 a_1^2 a_3^2 + 2 a_2^2 a_3 + 2 a_1 a_2 a_4 + a_1 a_5 &=& 0 \\
&\vdots&
\end{eqnarray*}$$

### Subcase $a_0 = 0, a_1 = 1$

We have $$\sum_{j \ge 2} a_j\left(x^j + \left(x + \sum_{i \ge 2} a_i x^i\right)^j\right) = 0$$ and it's easy to show by induction that $\forall j \ge 2: a_j = 0$. Uninteresting case.

### Subcase $a_0 = 0, a_1 = -1$

By substituting from the earlier identities into the later ones we can simplify further. Note that half of the identities become trivial. NB I've also assumed characteristic not equal to 2 and halved both sides of all of the following identities:

$$\begin{eqnarray*}
a_3 &=& -a_2^2 \\
a_5 &=& 2 a_2^4 - 3 a_2 a_4 \\
a_7 ​&=& -13 a_2^6 + 18 a_2^3 a_4 - 2 a_4^2 - 4 a_2 a_6 \\
a_9 &=& 145 a_2^8 - 221 a_2^5 a_4 + 50 a_2^2 a_4^2 + 35 a_2^3 a_6 - 5 a_4 a_6 - 5 a_2 a_8 \\
&\vdots&
\end{eqnarray*}$$

I haven't attempted to prove that the pattern of the even coefficients being free continues, but it seems plausible that there is a family of $|\mathbb{R}|^{|\mathbb{N}|}$ series, and that doesn't make me optimistic about a neat non-circular characterisation.