Every automorphism of the polynomial algebra in \(n\) variables over a field \(k\) induces a bijection of the set \(k^n\) of \(n\)-tuples and in this way the automorphism group maps in the group of all bijections of \(k^n\). It is easy to see that for infinite fields very few bijections are images of automorphisms. The paper under review studies the bijections of \({\mathbb F}_q^n\) which correspond to automorphisms of the polynomial algebra, where \({\mathbb F}_q\) is the field with \(q\) elements. The main result is surprising. If \(q\) is odd or \(q=2\), then every bijection is an image of an automorphism. If \(q=2^r\) and \(r>1\), then only the even bijections correspond to tame automorphisms. This looks quite perspective for checking various conjectures on automorphisms and constructing counterexamples over finite fields and, maybe also over arbitrary fields. As a consequence the author obtains that a set \(S\subset {\mathbb F}_q^n\) is the set of zeros of a coordinate (i.e. of an image of \(x_1\) under some automorphism of \({\mathbb F}_q[x_1,\ldots,x_n]\)) if and only if \(S\) has exactly \(q^{n-1}\) elements. The proof of the main result is based on explicit constructions and on the theorem of Jordan that any primitive subgroup of the symmetric group \({\mathcal S}_m\) which contains a 3-cycle contains also the alternating group \({\mathcal A}_m\).