Starting with $x$, which is a positive integer or zero, and $y$ a second positive integer or zero, with $y \ne x$, we can create lists.
Set $p$ a prime greater than 2, $\alpha = (p-1)/2$, and $\beta=\lfloor \log_2p \rfloor$. Then we can create 2 ordered sets; a set using $x$ and a set using $y$:
$$\left( x^\alpha, (x+a_1)^\alpha, (x+a_2)^\alpha, \dots (x+a_\beta)^\alpha \right) \pmod p$$ $$\left( y^\alpha, (y+a_1)^\alpha, (y+a_2)^\alpha, \dots (y+a_\beta)^\alpha \right) \pmod p$$
...where $a_k$ is a natural less than or equal to $p$. Note that unless $x$ or $y$ equals $-a_j$ for some $j$, each entry is either $1$ or $-1$ mod $p$ (depending on whether the number being exponentiated is a quadratic residue of nonresidue), so there are at most $2^{\beta+1}$ possible tuples not containing $0$; note also that $2^{\beta+1}$ is between $p$ and $2p$.
My question is: If we are trying to maximize the amount of unique sets, and we are allowed to choose the $a_k$'s any way we like in order to maximize the number of unique sets, how many unique sets can we get?
NOTES
This question is similar to "Is this sliding window unique?", except here we are allowed to pick $a_k$ values in order to try to maximize the number of unique sets and reduce the number of matching "tuples".