Let $P \in \mathbb{F}_q[x_1, \dots, x_n][y]$ be some fixed polynomial. Substituting polynomials for the variables in $P$ gives a map $\mathbb{F}_q[x]^{n+1} \to \mathbb{F}_q[x]$ defined by $$(Q_1(x), \dots, Q_n(x), Y(x)) \mapsto P(Q_1(x), \dots, Q_n(x))(Y(x)).$$
Suppose that for all $Y \in \mathbb{F}_q[x]$, the resulting polynomial $P(Q_i)(Y)$ is not the zero polynomial in $\mathbb{F}_q[x]$. Is this condition equivalent to some finite set of algebraic criteria on the $Q_i$?
The finite set of criteria should of course depend on $P$; it is also fine if the set depends on the characteristic $q$, so feel free to pick your favorite one and work there. For example, writing $P(Q_i)(Y) = c_d(Q_i)Y^d + \cdots + c_0(Q_i)$, it is necessary that $c_0(Q_i) \neq 0$ (take $Y$ to be the zero polynomial) and similarly that $\sum_{j=0}^d c_i(Q_i) \neq 0$ (take $Y = 1$). But these clearly are not sufficient.
I have some reason to suspect that there is no such finite set, but the argument sketch is very roundabout and there are some finicky details. I'd like to know if there is a more direct way to see this. It seems at first glance that a simple linear-algebra-style argument should work, but Lucas's theorem on binomial coefficients mod $q$ makes me suspicious that the details here are quite a bit trickier than one might expect.
I would also be interested in versions of this question where $\mathbb{F}_q$ is replaced by other fields or rings (preferably finite or countably infinite), especially if there is a positive answer to my question.
(This question is crossposted from Math.SE, where it was originally asked over a week ago but got no responses.)
