Let $F$ be the distribution supported on the numbers $x_1 \lt x_2 \lt \ldots \lt x_n$ that assigns probabilities $p_i \gt 0$ to each $x_i.$ By definition, its (raw) moment of degree $k$ is
$$\mu_k = \sum_{i=1}^n p_i x_i^k.$$
I will begin with a series of observations about this situation, each of interest in its own right. A basic tool is the sequence of vectors $\mathbf{x}_k = (x_1^k, x_2^k, \ldots, x_n^k)$ for $k=0, 1, \ldots,n-1.$ Writing $\mathbf{p} = (p_1,p_2,\ldots, p_n),$ each moment can be expressed as a vector product
$$\mu_k = \sum_{i=1}^n p_i x_i^k = \mathbf{p}\, \mathbf{x}_k^\prime.$$
The collection $\{\mathbf{x}_0,\mathbf{x}_1, \ldots, \mathbf{x}_{n-1}\}$ is linearly independent. To show this, assume the contrary: that is, let coefficients $c_k$ not all zero be such that $$\sum_{k=0}^{n-1} c_k \mathbf{x}_k = \mathbf{0}.\tag{1}$$ Written out component-by-component, $(1)$ asserts that for each $i=1,2,\ldots, n,$ $$\sum_{k=0}^{n-1} c_k x_i^k = 0.$$ That exhibits each $x_i$ as a root of the polynomial $c(T)=c_{n-1}T^{n-1}+c_{n-2}T^{n-2}+\cdots + c_0.$ Such a polynomial has at most $\operatorname{deg}(c)\le n-1$ distinct roots, contradicting the distinctness of the $n$ $x_i.$
All moments are determined by the first $n$ moments $\mu_0,\mu_1,\ldots,\mu_{n-1}.$ The preceding result shows the vectors $\mathcal{X} = \{\mathbf{x}_k,k=0,1,\ldots, n-1\},$ are a basis for $\mathbb{R}^n.$ Therefore for any $m,$ $\mathbf{x}_m$ is a linear combination of the $\mathbf{x}^k,$ $k=0,1,\ldots,n-1;$ that is, there exist coefficients $\,_ma_k$ (determined solely by the $x_i$) for which $$\mathbf{x}_m = \,_ma_0\mathbf{x}_0 + \,_ma_1\mathbf{x}_1 + \cdots + \,_ma_{n-1}\mathbf{x}_{n-1}.$$ Consequently $$\mu_m = \mathbf{p}\,\mathbf{x}_m^\prime = \mathbf{p}\,\sum_{i=0}^{n-1}\,_ma_k \mathbf{x}_k^\prime = \sum_{i=0}^{n-1}\,_ma_k \mathbf{p}\,\mathbf{x}_k^\prime= \sum_{i=0}^{n-1}\,_ma_k \mu_k.$$
The numbers $x_i$ and the first $n$ moments determine $\mathbf{p}.$ Indeed, the first $n$ moments are the coefficients of $\mathbf{p}$ in the basis dual to $\mathcal X.$
The first $n$ moments of $F$ determine, and are determined by, the distribution shifted by a constant $\lambda.$ This is the distribution supported on $x_1-\lambda, x_2-\lambda, \ldots, x_n-\lambda$ with probabilities $p_i.$ The demonstration is straightforward: use the Binomial theorem to expand $(x_i-\lambda)^k$ in terms of $x_i^0, x_i^1, \ldots, x_i^k.$
Part of the question is whether there exist $n^\prime,$ a positive probability vector $\mathbf{q},$ and support points $y_1\lt y_2\lt \ldots \lt y_{n^\prime},$ determining a distribution $G$ having the same moments as $F.$ Suppose there is. Shift both distributions by $\lambda=-\min(x_1,y_1),$ simplifying the situation to distributions with nonnegative support. By taking $m$ arbitrarily large, the largest support points eventually dominate the moments: $$q_{n^\prime} y_{n^\prime}^m \approx \mu_m \approx p_n x_n^m$$ This is possible only when $q_{n^\prime}=p_n$ and $y_{n^\prime} = x_n.$ Continuing inductively, we conclude $n=n^\prime,$ $\mathbf{q}=\mathbf{p},$ and $\mathbf{x}_1=\mathbf{y}_1:$ that is, $G=F.$
Finally, how many moments need to be known to determine $\mathbf{p}$ and $\mathbf{x}$? Consider the map $f:\mathbb{R}^n\times \mathbb{R}^n\approx \mathbb{R}^{2n}\to\mathbb{R}^{2n}$ defined by $$f(\mathbf{p}^\prime, \mathbf{x}^\prime) = (\mathbf{p}\mathbf{x}_0^\prime, \mathbf{p}\mathbf{x}_1^\prime, \ldots, \mathbf{p}\mathbf{x}_{2n-1}^\prime)^\prime.$$ Its derivative is the $2n\times 2n$ matrix
$$Df(\mathbf{p}^\prime, \mathbf{x}^\prime) = \pmatrix{1 & \cdots & 1 & 0 & \cdots & 0 \\
x_1 & \cdots & x_n & p_1 & \cdots & p_n \\
x_1^2 & \cdots & x_n^2 & 2p_1x_1 & \cdots & 2p_n x_n \\
\vdots & \cdots & \vdots & \vdots & \cdots & \vdots \\
x_1^{2n-1} & \cdots & x_n^{2n-1} & (2n-1)p_1x_1^{2n-2} & \cdots & (2n-1)p_nx_n^{2n-2}}$$
with a Vandermonde-like structure, enabling us to obtain a simple formula for its determinant,
$$\operatorname{Det}\left(Df(\mathbf{p}^\prime, \mathbf{x}^\prime)\right) = -(p_1p_2\cdots p_n)^{2n} \left(\prod_{1\le i\lt j \le n}(x_i-x_j)\right)^4.$$
Because none of the $p_i$ is zero and all the $x_i$ are distinct, this is nonzero. The Inverse Function Theorem implies $f$ is locally invertible: that is, provided $\mathbf{\mu}=(\mu_0,\mu_1,\ldots,\mu_{2n-1})$ is in the range of $f$, there exists an inverse $f^{-1}\subset\mathbb{R}^n\times \mathbb{R}^n$ in a neighborhood of $\mathbf{\mu}.$ That is,
The first $2n$ moments $\mu_0,\mu_1,\ldots,\mu_{2n-1}$ determine a discrete set of solutions $(\mathbf{p},\mathbf{x})$ corresponding to those moments.
As we have already shown, all such solutions correspond to the same distribution: they differ only by permuting the indexes $1,2,\ldots, n$ of the variables.
