[proofplan]
We use the field-theoretic norm to exploit the global Galois structure of the cyclotomic field $\mathbb{F} = \mathbb{Q}(\varepsilon)$, where $\varepsilon = e^{2\pi i / n}$. The norm $N(\alpha) = \prod_{\sigma \in \mathcal{G}} \sigma(\alpha)$ over $\mathcal{G} = \operatorname{Gal}(\mathbb{F}/\mathbb{Q})$ is rational (Galois-invariant) and an algebraic integer (product of Galois conjugates of an algebraic integer), hence an ordinary integer. Each Galois conjugate $\sigma(\alpha)$ is again an average of $m$ roots of unity (Galois automorphisms permute roots of unity), so the triangle inequality gives $|\sigma(\alpha)| \le 1$. Thus $|N(\alpha)| \le 1$, so $N(\alpha) \in \{-1, 0, 1\}$. The case $N(\alpha) = 0$ forces $\alpha = 0$, while $|N(\alpha)| = 1$ forces every $|\sigma(\alpha)| = 1$; in particular $|\alpha| = 1$.
[/proofplan]
[step:Place $\alpha$ inside the cyclotomic field $\mathbb{F} = \mathbb{Q}(\varepsilon)$]
Set $\varepsilon = e^{2\pi i / n} \in \mathbb{C}$, a primitive $n$-th root of unity, and let $\mathbb{F} = \mathbb{Q}(\varepsilon)$ be the $n$-th cyclotomic field. Every $n$-th root of unity is a power of $\varepsilon$ and therefore lies in $\mathbb{F}$. By hypothesis each $\lambda_j$ satisfies $\lambda_j^n = 1$, so $\lambda_j \in \mathbb{F}$ for all $j$. Since
\begin{align*}
\alpha = \frac{1}{m} \sum_{j=1}^m \lambda_j
\end{align*}
is a $\mathbb{Q}$-linear combination of elements of $\mathbb{F}$, we conclude $\alpha \in \mathbb{F}$.
The extension $\mathbb{F}/\mathbb{Q}$ is a finite Galois extension (the cyclotomic extension is the splitting field of $X^n - 1 \in \mathbb{Q}[X]$, which is separable since $\operatorname{char}\mathbb{Q} = 0$). Let $\mathcal{G} = \operatorname{Gal}(\mathbb{F}/\mathbb{Q})$.
[/step]
[step:Define the field norm $N(\alpha)$ and show it is a rational number]
Define the norm
\begin{align*}
N: \mathbb{F} &\to \mathbb{F} \\
\beta &\mapsto \prod_{\sigma \in \mathcal{G}} \sigma(\beta).
\end{align*}
We claim $N(\alpha) \in \mathbb{Q}$. For any $\tau \in \mathcal{G}$,
\begin{align*}
\tau(N(\alpha)) = \tau\!\left( \prod_{\sigma \in \mathcal{G}} \sigma(\alpha) \right) = \prod_{\sigma \in \mathcal{G}} (\tau \sigma)(\alpha) = \prod_{\sigma' \in \mathcal{G}} \sigma'(\alpha) = N(\alpha),
\end{align*}
where in the second equality we used that $\tau$ is a field automorphism (preserves products), and in the third we used that left multiplication $\sigma \mapsto \tau\sigma$ is a bijection of $\mathcal{G}$.
Thus $N(\alpha)$ is fixed by every element of $\mathcal{G}$, i.e.\ $N(\alpha) \in \mathbb{F}^{\mathcal{G}}$. By the fundamental theorem of Galois theory applied to the Galois extension $\mathbb{F}/\mathbb{Q}$, the fixed field of $\mathcal{G}$ equals $\mathbb{Q}$. Therefore $N(\alpha) \in \mathbb{Q}$.
[/step]
[step:Show $N(\alpha)$ is an algebraic integer, hence in $\mathbb{Z}$]
Each $\lambda_j$ is a root of $X^n - 1 \in \mathbb{Z}[X]$, so each $\lambda_j$ is an algebraic integer. Therefore the $\mathbb{Z}$-linear combination
\begin{align*}
m \alpha = \sum_{j=1}^m \lambda_j
\end{align*}
is an algebraic integer (algebraic integers form a ring). By hypothesis $\alpha$ itself is an algebraic integer.
Algebraic integers are closed under Galois conjugation: if $\beta$ is an algebraic integer and $\sigma \in \mathcal{G}$, then applying $\sigma$ to a monic integer polynomial relation $\beta^d + c_{d-1}\beta^{d-1} + \cdots + c_0 = 0$ (with $c_i \in \mathbb{Z}$ fixed by every field automorphism) gives the same relation for $\sigma(\beta)$. Hence each $\sigma(\alpha)$ is an algebraic integer, and the product
\begin{align*}
N(\alpha) = \prod_{\sigma \in \mathcal{G}} \sigma(\alpha)
\end{align*}
is also an algebraic integer.
We now use the standard fact: any rational number that is also an algebraic integer lies in $\mathbb{Z}$. (If $p/q$ in lowest terms satisfies a monic integer polynomial of degree $d$, clearing denominators gives $p^d + c_{d-1}p^{d-1}q + \cdots + c_0 q^d = 0$, so $q \mid p^d$; coprimality forces $q = \pm 1$.) Applying this to $N(\alpha) \in \mathbb{Q}$, we conclude
\begin{align*}
N(\alpha) \in \mathbb{Z}.
\end{align*}
[/step]
[step:Bound $|\sigma(\alpha)| \le 1$ for every $\sigma \in \mathcal{G}$]
Fix $\sigma \in \mathcal{G}$. The automorphism $\sigma$ fixes $\mathbb{Q}$ pointwise, so
\begin{align*}
\sigma(\alpha) = \sigma\!\left( \frac{1}{m} \sum_{j=1}^m \lambda_j \right) = \frac{1}{m} \sum_{j=1}^m \sigma(\lambda_j).
\end{align*}
Each $\sigma(\lambda_j)$ is again an $n$-th root of unity: $\sigma(\lambda_j)^n = \sigma(\lambda_j^n) = \sigma(1) = 1$. In particular $|\sigma(\lambda_j)| = 1$. Applying the triangle inequality,
\begin{align*}
|\sigma(\alpha)| = \left| \frac{1}{m} \sum_{j=1}^m \sigma(\lambda_j) \right| \le \frac{1}{m} \sum_{j=1}^m |\sigma(\lambda_j)| = \frac{1}{m} \cdot m = 1.
\end{align*}
[/step]
[step:Combine $|N(\alpha)| \le 1$ with $N(\alpha) \in \mathbb{Z}$ to conclude]
From Step 4, $|\sigma(\alpha)| \le 1$ for every $\sigma \in \mathcal{G}$, so
\begin{align*}
|N(\alpha)| = \prod_{\sigma \in \mathcal{G}} |\sigma(\alpha)| \le 1.
\end{align*}
Combined with $N(\alpha) \in \mathbb{Z}$ (Step 3), we have $N(\alpha) \in \{-1, 0, 1\}$.
**Case $N(\alpha) = 0$.** Some factor $\sigma_0(\alpha)$ vanishes; since $\sigma_0$ is a field automorphism (in particular injective), $\alpha = 0$.
**Case $|N(\alpha)| = 1$.** We have $\prod_{\sigma} |\sigma(\alpha)| = 1$ with each $|\sigma(\alpha)| \le 1$. Suppose some $|\sigma_1(\alpha)| < 1$; then since $|\sigma(\alpha)| \le 1$ for the remaining factors, the product would be strictly less than $1$, contradicting $|N(\alpha)| = 1$. Hence $|\sigma(\alpha)| = 1$ for every $\sigma \in \mathcal{G}$. Taking $\sigma = \operatorname{id}_{\mathbb{F}}$ in particular gives $|\alpha| = 1$.
In either case, $\alpha = 0$ or $|\alpha| = 1$, as claimed.
[guided]
The strategy is to leverage the rigidity of integers: there are only finitely many integers in any bounded interval. A complex number with no special structure can be very small in absolute value without being zero. But an *algebraic integer* in $\mathbb{Q}$ is forced to be an ordinary integer, and the only integer in $[-1, 1]$ with $|\cdot| < 1$ is $0$. So if we can show the rational integer $N(\alpha)$ has absolute value $\le 1$, then $N(\alpha) \in \{-1, 0, 1\}$ — a discrete finite list.
How do we get from $\alpha$ (which is *not* in $\mathbb{Q}$ in general) to something rational? The Galois norm $N(\alpha) = \prod_\sigma \sigma(\alpha)$ is the standard tool. It is rational because it is Galois-invariant (the product is a symmetric function of the Galois orbit), and it is an algebraic integer because each factor is. Combining these two facts forces $N(\alpha) \in \mathbb{Z}$.
The bound $|\sigma(\alpha)| \le 1$ is where the special hypothesis on $\alpha$ enters: the formula $\alpha = (1/m) \sum_j \lambda_j$ shows $\alpha$ is an *average* of unit-modulus complex numbers, so $|\alpha| \le 1$ by the triangle inequality. The crucial observation is that the same bound holds for every Galois conjugate $\sigma(\alpha)$, because $\sigma$ permutes the roots of unity — applying $\sigma$ to the formula gives $\sigma(\alpha) = (1/m)\sum_j \sigma(\lambda_j)$, where each $\sigma(\lambda_j)$ is *still* a root of unity (just possibly a different one).
Multiplying the unit-bounds across the entire Galois orbit gives $|N(\alpha)| \le 1$. Discreteness ($N(\alpha) \in \mathbb{Z}$) closes the argument. The dichotomy at the end ("either some factor is zero, or all factors have modulus $1$") is the standard observation that a product of complex numbers of modulus $\le 1$ has modulus $1$ if and only if every factor has modulus $1$.
[/guided]
[/step]
