Multiplicative self-maps of a finite field ($f(xy)=f(x)f(y)$)

For a finite field with $q$ elements, exactly $q+1$ functions from the field to itself preserve multiplication, without requiring additivity or preservation of the identity. They are the constant-zero map, the constant-one map, and the $q-1$ power maps $x\mapsto x^k$ for $1\le k\le q-1$.

To prove completeness, $f(1{)}^2=f(1)$ forces $f(1)\in \{0,1\}$ because a field has no zero divisors. The zero choice forces the constant-zero map. For $f(1)=1$, nonzero elements must map to nonzero elements, and the restriction is a group homomorphism of the cyclic multiplicative group of a finite field. Its generator has $q-1$ possible images, giving the $q-1$ power maps on nonzero elements. Setting $f(0)=0$ extends each. The only further possibility is $f(0)=1$, which forces every value to be one. For $q=5$, this gives six maps. These are generally not field homomorphisms, because preservation of addition was never required.