For a non-constant polynomial $A \in \mathbb{Z}[x]$, let $\mathcal{P}(A)$ denote the set of prime numbers $p$ which divide $A(n)$ for some integer $n$. If $\mathcal{P}(A) \subseteq \mathcal{P}(B)$ for some $A,B$, does there necessarily exist $C \in \mathbb{Q}[x]$ such that $A|B\circ C$? (Here, $B \circ C = B(C(x))$ is the usual polynomial composition). If so, is there an efficient way to find such $C$? This would provide a nice method to check, for example, if two polynomials have the same set of prime factors: if $A(x) = x^3 - 7x -7$ and $B(x) = x^3 + x^2 - 2x - 1$, we can take $C(x) = x^2 - x - 5$, and if we swap $A,B$, $C(x) = x^2 + 2x - 1$ works, proving that $\mathcal{P}(A) = \mathcal{P}(B).$ The question is whether there always exists such $C$. We can ask more: can we conclude from $\mathcal{P}(A) \subseteq \mathcal{P}(B)$ that the splitting field of $A$ contains that of $B$ (over $\mathbb{Q}$)?

If the answer is negative, is there any nice way to characterize the counterexamples, or to find one with $A$ of minimal degree?