Can this system of polynomial equations be solved in polynomial time?

I have these $n$ equations, with $n$ variables. Variables are first $n$ positive integers, constants can be any rational number including zero. Given that there is always a solution, how do we find a solution to the system:

\begin{align}
ax^{1}+by^{1}+cz^{1}+\cdots&=k_{1}\\
ax^{2}+by^{2}+cz^{2}+\cdots&=k_{2}\\
ax^{3}+by^{3}+cz^{3}+\cdots&=k_{3}
\end{align}

so on ... till
$$ax^{n}+by^{n}+cz^{n}+\cdots=k_{n}$$

for variables $x,y,z\in \{1,2\ldots, n\}$.

The value of constants $a$,$b$,$c$... remain same in all these equations.

And these constants can also be equal to each other. For e.g: $a = b$ so that values of $x$ and $y$ can become interchangeable. In such cases one working solution is suffice.

Can we find a solution or determine none exist in polynomial time?

Edit: For the sake of clarity here is a simple example.

If $n=3$, I know $x,y,z$ can only take 1, 2 or 3 as their values. So if I have $x + y + z = 9$, I clearly know that $x, y$ and $z$ are all 3. But if I have $x + y + z = 6$, I don't know whether they are $2,2,2$ or $1,2,3$. Then I can make use of $x^2 + y^2 + z^2 = 14$. Then I will know they are $1,2,3$.

Let us denote the unknowns by $x_1,\ldots,x_n$, and the coefficients by $c_1,\ldots,c_n$. For a polynomial $P(x)$, we denote
$$
[P(x)] = \sum_{i=1}^n c_i P(x_i).
$$
We are given the values of $[1],[x],\ldots,[x^n]$, from which we can calculate $[P(x)]$ for every polynomial of degree at most $n$. In particular, we can compute, for each $j \in \{1,\ldots,n\}$, the value
$$
\left[ \prod_{k \neq j} \frac{x-k}{j-k} \right] = \sum_{i\colon x_i=j} c_i.
$$
To complete the solution, we need to solve several SUBSET-SUM instances, which possibly interact.

Conversely, we can encode SUBSET-SUM using your equations. Given a set of positive coefficients $c_1,\ldots,c_n$ summing to $C$ and a target $S$, we encode a solution using $\{1,2\}$-valued variables (where $1$ represents a value in the subset summing to $S$), and can compute $[x^k] = S + (C-S)2^k$. These numbers satisfy
$$
[x^k(x-1)(x-2)] = [x^{k+2}-3x^{k+1}+2x^k] = \\
(S + (C-S)2^{k+2}) - 3(S+(C-S)2^{k+1}) + 2(S+(C-S)2^k) = \\
(1-3+2)S + (4-6+2)(C-S)2^k = 0.
$$
Therefore the argument above shows that every $\{1,\ldots,n\}$-valued solution is actually $\{1,2\}$-valued. Therefore a solution to your problem exists if and only if the SUBSET-SUM instance is a Yes instance.

When all coefficients $c_i$ are equal, the argument above allows us to calculate directly the number of $x_i$'s equal to $j$. Let us take your example: $[1] = 3$, $[x] = 6$ and $[x^2] = 14$. Denoting by $y_j$ the number of $x_i$'s equal to $j$, we compute
$$
\begin{align*}
y_1 &= \left[ \frac{(x-2)(x-3)}{(1-2)(1-3)} \right] = \frac{[x^2-5x+6]}{2} = \frac{14-5\cdot6+6\cdot3}{2} = 1, \\
y_2 &= \left[ \frac{(x-1)(x-3)}{(2-1)(2-3)} \right] = [-x^2+4x-3] =
-14+4\cdot6-3\cdot3 = 1, \\
y_3 &= \left[ \frac{(x-1)(x-2)}{(3-1)(3-2)} \right] = \frac{[x^2-3x+2]}{2} =
\frac{14-3\cdot6+2\cdot3}{2} = 1.
\end{align*}
$$

StackExchange Computer Science Q#105572, answer score: 3