First of all, thank you Gerry Myerson for bringing this problem to my attention at West Coast Number Theory 2019. 

The answer to this question **seems to be No**.

**Edit on 12/25** : There was my misunderstanding of the paper by Lam and Leung at Step 3, pointed by ***bold face italic***.

We prove that under the assumptions of this problem, $|X|$ must be prime. 

**Step 1 : Reduction**

Let $|X|>2$ and $\zeta_N=\exp(2\pi i (1/N))$. 

The sequence $s_n=\sum_{x\in X} x^n$ satisfies a linear recurrence relation. By [Skolem-Mahler-Lech theorem](https://en.wikipedia.org/wiki/Skolem%E2%80%93Mahler%E2%80%93Lech_theorem), there is an arithmetic progression $\{an+b\}_{n\geq 0}\subseteq \mathbb{N}$ with $a>0$, $b\in \mathbb{N}$ such that 
$$
s_{an+b}=\sum_{x\in X} x^{an+b} = 0 \ \textrm{for all} \ n\in\mathbb{N}.
$$

**Step 2 : Vandermonde**

Let $X=\{x_1,\ldots,x_k\}$. The result from Step 1 forms a Vandermonde system
$$
\begin{pmatrix} 1 & 1 & \cdots & 1 \\ x_1^a & x_2^a & \cdots & x_k^a \\ x_1^{2a} & x_2^{2a} & \cdots & x_k^{2a} \\
\vdots& \ddots& \cdots& \vdots \\
x_1^{(k-1)a} & x_2^{(k-1)a} & \cdots & x_k^{(k-1)a}\end{pmatrix}\begin{pmatrix} x_1^b \\ x_2^b \\ \vdots \\ x_k^b \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}.
$$
Since $x_i\neq 0$, the Vandermonde matrix must be singular. This yields $x_i^a = x_j^a$ for some $i\neq j$. Without loss of generality, assume that $x_1^a=x_2^a$. Then we may rewrite the system as 
$$
\begin{pmatrix} 1 & 1 & \cdots & 1 \\ x_2^a & x_3^a & \cdots & x_k^a \\ x_2^{2a} & x_3^{2a} & \cdots & x_k^{2a} \\
\vdots& \ddots& \cdots& \vdots \\
x_2^{(k-2)a} & x_3^{(k-2)a} & \cdots & x_k^{(k-2)a}\end{pmatrix}\begin{pmatrix} x_1^b+x_2^b \\ x_3^b \\ \vdots \\ x_k^b \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}.
$$
If $x_1^b+x_2^b=0$, it yields a shorter vanishing sums in the expression of $s_{an+b}=0$. So, we must have $x_1^b+x_2^b\neq 0$. Thus the above Vandermonde matrix is also singular, and we obtain another $i\neq j$ (both $\geq 2$) with $x_i^a=x_j^a$. Repeating this process, we obtain 
$$
x_1^a = x_2^a = \cdots = x_k^a. 
$$
Dividing by $x_1$, we may assume that all members of $X$ are roots of unity. 

**Step 3 : Vanishing sums of roots of unity**

We refer to the results of this paper: T. Y. Lam, K. H. Leung, ['On the vanishing sums of roots of unity'](https://www.sciencedirect.com/science/article/pii/S0021869399980894?via%3Dihub)

In this paper, the vanishing sums of roots of unity $\alpha_1+\cdots+\alpha_k=0$ is called *minimal* if no proper sums vanish. The characterization of the minimal vanishing sums of roots of unity is

 

(1) $1+ \zeta_p+ \cdots +\zeta_p^{p-1}=0$ for prime $p$, or


(2) $(\zeta_p+\cdots + \zeta_p^{p-1})(\zeta_q+\cdots+\zeta_q^{q-1})+(\zeta_r+\cdots+\zeta_r^{r-1}) = (-1)(-1)-1= 0$ for primes $p<q<r$.

and these forms multiplied by a root of unity $\zeta_s^t$ ..***or many more..***


If we have (1), then $|X|=p$ is prime. 

If we have (2), then consider $pq$-th powers. The sum in (2) after $pq$-th powers of each term, we obtain
$$
(1+\cdots +1) ( 1+ \cdots +1) + (\zeta_r^{pq} + \cdots + \zeta_r^{pq(r-1)})
$$
Since $(pq,r)=1$, the sum $(\zeta_r^{pq} + \cdots + \zeta_r^{pq(r-1)})$ is a rearrangement of $\zeta_r+ \cdots + \zeta_r^{r-1}$. This yields a vanishing sub-sum of length $r$, 
$$
1+\zeta_r+ \cdots + \zeta_r^{r-1}=0.
$$
Note that $(p-1)(q-1)+r-1 > r$. Thus, we cannot have (2) under the inclusion-minimal assumption of this problem.