Let $n \in \mathbb Z$ with $n \ge 3$ and let $p$ be a prime number such that $n|p-1$. Let $a_1,a_2,\dots,a_{2n-1} \in \mathbb Z/p\mathbb Z$. Suppose that the same class is represented by at most $n-1$ elements in the sequence of $a_i$'s. Since $\frac{2n-1}{n-1} > 2$, there are at least $3$ distinct classes in the sequence $a_1,\dots,a_{2n-1}$.

Let $s$ such that $ord_p(s) = n$ and consider $b_1s^{n-1} + b_2s^{n-2} + \dots + b_{n-1}s + b_n \pmod p$. A "valid operation" is to choose $n$ numbers among $a_1,\dots,a_{2n-1}$ and replace them in place of $b_1, \dots, b_n$ in some order.

It's easy to show that the cyclic permutations reach only the class $0 \pmod p$ or reach $n$ distinct classes $\pmod p$, because if $1 \le l < k \le n$ then: $$b_{k+1}s^{n-1} + \dots + b_ns^k + b_1s^{k-1} + \dots + b_k \equiv b_{l+1}s^{n-1} + \dots + b_ns^l + b_1s^{l-1} + \dots + b_l \pmod p$$ $$\iff (s^{n-k-2} - s^{n-l-2})(b_1s^{n-1} + \dots + b_n) \equiv 0 \pmod p$$ $$\iff l = k \text{ (contradiction, so all cyclic permutations are distinct) or }$$ $$b_1s^{n-1} + \dots + b_n \equiv 0 \pmod p \Rightarrow b_{k+1}s^{n-1} + \dots + b_ns^k + b_1s^{k-1} + \dots + b_k \equiv 0 \pmod p,$$ so all cyclic permutations reach $0 \pmod p$.

Question: Is it possible to show that valid operations reach at least $n+1$ classes $\pmod p$? I just need one more element.

For example, if $n = 3$ and $p = 7$ then the possibilities for $a_1,\dots,a_5$ are the following: $a,a,b,b,c$; $a,a,b,c,d$ and $a,b,c,d,e$ (where different letters represent different classes $\pmod p$).

I can answer this question if $n=3$. In fact, let $p \equiv 1 \pmod 3$ and consider $ord_p(s) = 3$. Select $a,b,c$ distinct classes in the sequence of $5$ elements. We only need to show that $as^2 + cs + b \not\in \{as^2 + bs + c, bs^2 + cs + a, cs^2 + as + b\}$ as classes $\pmod p$, and this is just an easy calculation (we use $s \not\in \{-1,0,1\} \pmod p$).

Perhaps Cauchy-Davenport inequality will help us. Thanks!