Qn: Let $p$ be an odd prime. For $x_1,\dots,x_p\in \Bbb Z, 0\le x_1\le x_2\le\cdots\le x_p\le p-1$, let $S_n=x_1^n+\cdots+x_p^n$, $0\le n\le p-2$.
Find the number of set of solutions $y_1,\dots,y_p\in \Bbb Z,0\le y_1\le y_2\le\cdots\le y_p\le p-1$ such that $$y_1^n+\cdots+y_p^n\equiv S_n\pmod p,\quad \forall 0\le n\le p-2.$$
Attempt
Let \begin{align*} f(t)&=(t-x_1)\cdots(t-x_p)=a_p t^p+a_{p-1} t^{p-1}+\cdots +a_0,\\ g(t)&=(t-y_1)\cdots(t-y_p)=b_p t^p+b_{p-1} t^{p-1}+\cdots +b_0. \end{align*} By considering Newton's identities, we could get $$f(t)\equiv g(t)\pmod p$$ or $$a_i\equiv b_i\pmod p,\quad \forall 0\le i \le p.$$
I think the solution is unique, so I tried to prove $$\frac{a_i-b_i}{a_j-b_j}=C,\quad \forall 0\le i,j \le p$$ and got stuck here.
Edit
The solution is not unique, for the case $p=5$, let $S_n=1^n+2^n+3^n+4^n+5^n$. We have $$1^n+1^n+1^n+1^n+1^n\equiv S_n\pmod p,\quad \forall 0\le n\le p-2,$$ where $\{1,1,1,1,1\}\neq\{1,2,3,4,5\}$.
Hint; (I assume you have $n \geq 1$) Given a monic Polynomial of degree $k$ namely $f(t)$ that splits entirely with roots $\alpha_1, \alpha_2...\alpha_k$, there is a unique monic polynomial of degree $k$; $g(t)$ with roots $\beta_1,...\beta_k$ such that
(1) $\alpha_1+\alpha_2...+\alpha_k = \beta_1+\beta_2...+\beta_k$
(2) $\alpha_1^2+\alpha_2^2...+\alpha_k^2 = \beta_1^2+\beta_2^2...+\beta_k^2$
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(k-1) $\alpha_1^{k-1}+...\alpha_{k}^{k-1} = \beta_1^{k-1}+\beta_2^{k-1}...+\beta_k^{k-1}$
which satisfies $f(t) = g(t)$. Now if $f(t) = g(t)$ then $\{\alpha_1,\alpha_2...\alpha_k\}$ = $\{\beta_1,\beta_2...\beta_k\}$ applying this idea you get that there is a unique solution. For the crux of the proof you would be able to see uniqueness of the polynomial by observing this wikipedia page
https://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial