Let $Q\in \mathbb{Z}[x]$ be a polynomial defining an injective function $\mathbb{Z}\to\mathbb{Z}$. Does it define an injective function $\mathbb{Z}/p\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$ for some prime $p$?
2 Answers
Consider $Q(x)=x(2x1)(3x1)$. This gives an injective map $\mathbb Z\to \mathbb Z$, because $n<m \implies Q(n)<Q(m)$. However, this $Q$ is not injective over $\mathbb Z/p\mathbb Z$ for any $p$ because $Q(x)=0$ has three solutions when $p\geq 5$ and two solutions when $p\in \{2,3\}$.
While the original question has been answered, there is a beautiful result of Fried in On a conjecture of Schur which is relevant to such questions. Suppose $Q$ is a polynomial in ${\Bbb Q}[x]$ such that for infinitely many primes the induced map from ${\Bbb Z}/p{\Bbb Z}$ to ${\Bbb Z}/p{\Bbb Z}$ is bijective. Then $Q$ must be of the form
(i) $Q(x) = ax^n + b$,
or
(ii) $Q(x) = T_n(x)$, where $T_n(x)$ denotes the $n$th Chebyshev polynomial,
or
(iii) Compositions of functions of this type.
This established an old conjecture of Schur. See also the account Turnwald  On Schur's conjecture, which discusses the history of the problem, and gives a detailed proof, correcting inaccuracies in the earlier literature.

1$\begingroup$ how does a polynomial with rational coefficients induce a map $\mathbb{Z}/p\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$? $\endgroup$– user158636Aug 9, 2020 at 19:29

2$\begingroup$ Just take the primes not dividing denominators of coefficients and reduce. $\endgroup$– LuciaAug 9, 2020 at 19:30

1$\begingroup$ That is to say, part of the quantification is "for infinitely many primes $p$, there is an induced selfmap of $\mathbb Z/p\mathbb Z$ and it is bijective". (Of course, the first condition throws away only finitely many primes.) $\endgroup$– LSpiceAug 9, 2020 at 19:53