Let $f(x) = \prod_{i=0}^n(1+a_ix), a_i \neq 0$, and $C(f(x)): = (c_0, c_i, \cdots, c_n) $, where $f(x) = \sum_{i=0}^n c_i x^i$.
For a given monomial $g(x) = (1+mx)(1+nx)$, i'm interested in a quantity $\langle C(g), \frac{1}{C((1+x)^2)} \rangle$. Here $\langle, \rangle$ is the inner product and $\frac{1}{.}$ is element-wise reciprocal.
And my question is, is there any non-trivial mapping$(m_1, m_2 \neq 1)$, $h: m \to (m_1, m_2)$, such that
$\langle C((1+mx)(1+nx)), \frac{1}{C((1+x)^2} \rangle = \langle C((1+m_1x)(1+m_2x)(1+nx)), \frac{1}{C((1+x)^3} \rangle$ ?
So you want $$1 + \frac{m+n}2 +mn = 1 + \frac{m_1 + m_2 +n}3 + \frac{m_1m_2 + m_1n + m_2n}3 + m_1m_2n\,?$$ You can solve that for $m_2$ in terms of the other variables $$m_2 = \frac{3m + n + 6mn - 2m_1 - 2m_1n}{2 + 2m_1 + 2n + 6m_1n}$$
So you can choose whatever function $m \mapsto m_1$ you want, and find the corresponding $m_2$ per the equation to build $h$. If $m_1$ is not a trivial function of $m$, neither will $h$ be trivial. As Boxwood already noted, it isn't going to be independent of $n$.
A similar calculation can be done for any degree $g$.