This isn't quite right. Modulo an odd prime, x->x+1 is an even permutation. x -> 2/x might be odd or even: Mostly it's pairs, but if the field contains sqrt2 (and -sqrt2), these will be singletons. [Also assume we map 0 -> 0.] sqrt2 exists iff P = 8K+-1. The number of pairs is about (P-1)/2, but we need to subtract out any singletons. If the number of pairs remaining is even, then x -> 2/x is also an even permutation, so both of our generators are even, and we can't generate the the odd half of the group. P=5 and P=7 are both in this class. For P=11, there's no sqrt2, and there are five swaps, so this at least generates some odd permutations. For P=17, sqrt2 = 6 (and -6), so there are seven swaps, and 2/x is an odd permutation. Rich ----
We know for sure that x->x+1 and x->2/x do generate ALL (and just) the perm polys (proved by someone else & verified by my Maxima code), ....