Permutations of $(Z/pZ)^*$ 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 $b^{-1}$. 
 A: A similar concept is an orthomorphism of a group $G$. This is an automorphism $\theta: G \rightarrow G$ with the property that $g^{-1}\theta(g)$ is a bijection (equivalently an automorphism). Two orthomorphisms $\theta$, $\eta$ are orthogonal if $\theta^{-1} \eta$ is an orthomorphism.
A set of $k$ orthogonal orthomorphisms correspond to a set of $k$ mutually orthogonal latin squares with specified symmetries. In particular, the examples you give above are the prototypical examples of orthogonal orthomorphisms, and they give a set of $p-1$ MOLS of order $p$. From these one can easily construct the (desarguesian) projective plane of order $p$.
It seems to me that your question relaxes the condition that the orthomorphisms be automorphisms of $G$: you simply want functions. The relation to mutually orthogonal latin squares should still hold however. So you are essentially looking for a non-desarguesian projective plane of order $p$. As far as I know this problem is open, though none are known to exist. (And people have looked.)
Non-desguesian projective planes exist at prime power orders - so I guess that there will be inequivalent sets of functions with the properties you desire there.
