Let $\F_q$ be a finite field. For any function $f:\F_q\to \F_q$, there exists a polynomial $p\in \F_q[x]$ such that $f(a) = p(a)$ for all $a\in \F_q$. In other words, every function from a finite field to itself can be represented by a polynomial.

In particular, every permutation of $\F_q$ can be represented by a polynomial. This isn’t true for other finite rings, a fact which is the topic of one of my papers. At any rate, polynomials that represent permutations are called permutation polynomials.

It’s not easy to determine which polynomials are permutation polynomials. The exception is for the finite field $\F_2$, and more generally $\Z/2^w$. Then this case Ronald Rivest gave a straightforward criterion for whether a polynomial is a permutation polynomial.

However, there is a classical result that says any permutation may be represented uniquely by a polynomial whose degree is at most $q-2$. Here is an example noted by Charles Wells. Let $a,b\in\F_q$ be two distinct elements. The polynomial
$$f(x) = x + (a-b)(x-a)^{q-1} + (b-a)(x-b)^{q-1}$$
represents the transposition of $a$ and $b$. I suggest that you verify this by direct substitution, using the identity $c^{q-1} =1$ whenever $c\not=0$, which in turn follows from Fermat’s little theorem.
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