Let $p$ be a prime integer, and let $(\mathbb Z/p\mathbb Z)^*$ be the set of non-zero elements of $\mathbb Z/p \mathbb Z$. Denote by $S((\mathbb Z/p \mathbb Z)^*)$ the group of permutations of $(\mathbb Z/p \mathbb Z)^*$.

Say that a map $a:(\mathbb Z/p \mathbb Z)^*\to S((\mathbb Z/p \mathbb Z)^*)$ satisfies condition (A) if, for any two distinct elements $i,j\in (\mathbb Z/p \mathbb Z)^*$, $a(i)-a(j)\in S((\mathbb Z/p \mathbb Z)^*)$.

For example, let $a(i)(k) = ik.$ This satisfies condition (A). The same is true if we permute the functions $a'(i) = a(c(i))$, or relabel the objects $a''(i)(k) = i \cdot b(k)$, or both. Are these modifications of $a(i)(k) = ik$ the only ways to get a map satisfying condition (A)?

If $a$ satisfies (A), are there $b,c\in S((\mathbb Z/p \mathbb Z)^*)$ such that, for all $i\in (\mathbb Z/p \mathbb Z)^*$ and all $k\in (\mathbb Z/p \mathbb Z)^*$, $a(i)(k)=c(i)\cdot b(k)$, where the dot is multiplication in $\mathbb Z/p \mathbb Z$?

Note: it would probably be sufficient to prove that, if $a$ satisfies (A), then, for all $i,j,l\in (\mathbb Z/p \mathbb Z)^*$, $a(i)a(l)^{-1}a(j)=a(j)a(l)^{-1}a(i)$. Or in simpler terms, if $a(1)$ is the identity (one can reduces to this case) then the $a(i)$ commute.

**edit** I've corrected the question -- and the paragraph before it -- thanks to comments by François Brunault and Victor Protsak, who noted that the original formulation was incorrect due to an irrelevant $b^{-1}$.

notpreserve condition (A)! Keep in mind that a permutation $b$ is not, in general, compatible with the additive structure on $\mathbb{Z}/p\mathbb{Z}$ involved in checking the condition. For a specific example, consider $p=5$ and $b=(12).$ Then $a''(i)=b^{-1}(i\cdot b(k))$ are the following permutations: $$ i=1: 1\to 1, 2\to 2, 3\to 3, 4\to 4\qquad i=2: 1\to 4, 2\to 1, 3\to 2, 4\to 3 $$ Clearly, $a''(1)-a''(2)$ is not a permutation, since it maps $2,3,$ and $4$ to $1.$ $\endgroup$ – Victor Protsak May 21 '13 at 22:11