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, $\alpha = \lfloor p/2 \rfloor$, and $\beta=\lceil \log_2{(p)} \rceil$. Then we can create 2 ordered sets; a set using $x$ and a set using $y$:
$$\left( x^\alpha, (x+1)^\alpha, (x+2)^\alpha, \dots (x+\beta)^\alpha \right)$$ $$\left( y^\alpha, (y+1)^\alpha, (y+2)^\alpha, \dots (y+\beta)^\alpha \right)$$
My question is: If we find the values contained in these ordered sets, are they always unique modulo $p$? In other words, must the sets created using $x$ and $y$ be different modulo $p$?
No. For example if $x=7,y=11,p=23,\alpha=11,\beta=5$ $$ \begin{array}{lll} 7^\alpha & \equiv 11^\alpha &\equiv -1 \pmod{23} \\ 8^\alpha & \equiv 12^\alpha &\equiv +1 \pmod{23} \\ 9^\alpha & \equiv 13^\alpha &\equiv +1 \pmod{23} \\ 10^\alpha & \equiv 14^\alpha &\equiv -1 \pmod{23} \\ 11^\alpha & \equiv 15^\alpha &\equiv -1 \pmod{23} \\ 12^\alpha & \equiv 16^\alpha &\equiv +1 \pmod{23} \\ \end{array} $$
Since $\alpha=\lfloor p/2\rfloor = (p-1)/2$, for $n\not \equiv 0 \pmod p$ we have $n^\alpha \equiv \pm 1 \pmod{p}$. So for a choice of $x$ with $0<x<p-\beta$ there are only $2^{\beta+1}$ possible $(\beta+1)$-tuples that can result.
If the distribution of quadratic residues modulo $p$ behaves sufficiently like a random sequence of choices from $\{-1,+1\}$, and we choose $(x,y)$ with $0<x<y<p-\beta$, then to a first-order approximation there are about $p^2/2$ choices of $(x,y)$ and the tuples they produce match with probability $2^{-\beta-1}$. So for large $p$ we would expect somewhere between $p/16$ and $p/4$ choices of $(x,y)$ to match.
Here's a scatter plot of the number of pairs $(x,y)$ that produce matching tuples for the first 70 odd primes.
For example, for $p=199$ we get matches for $(x,y)=(1,25),(3,110),(4,111),(12,176)$ and fourteen others. Probably $p=67$ is the largest prime for which the $(\beta+1)$-tuples for $x$ and $y$ are different for every choice of $x\not\equiv y\pmod p$.