A similar fact is well known: A polynomial $f(x)$ with complex coefficients divides a polynomial with positive coefficients if and only $f(x)$ has no nonnegative root. A similar strategy works here.

Surely, we may work with polynomials with real coefficients, and we will do that (approximation works).

Say that a polynomial is $N$-extremal (for $N>1$) if $|a_i|> N\sum_{j\neq i}|a_j|$ for some $i$.

**Lemma.** The product of two $N$-extremal polynomials is $\frac{N-1}2$-extremal.

*Proof.* We may assume that
$$
N=|a_i|>N\sum_{j\neq i}|a_j| \quad\text{and}\quad
N=|b_s|>N\sum_{t\neq s}|b_t|
$$
for the coefficients of the two factors.
If the $c_k$'s are the product's coefficients, then
$$
|c_{i+s}|\geq |a_i||b_s|-\sum_{j\neq i}\sum_{t\neq s}|a_j||b_t|>N^2-1,
$$
while
$$
\sum_{k\neq i+s}|c_k|\leq |a_i|\sum_{t\neq s}|b_t|+|b_s|\sum_{j\neq i}|a_j|+\sum_{j\neq i}|a_j|\sum_{t\neq s}|b_t|< 2N+1,
$$
and $\frac{N^2-1}{2N+1}>\frac{N-1}2$. $\square$

Now consider a (say, monic) polynomial $f(x)$ with no roots on the unit circle. Expand it as
$$
f(x)=\prod_i(x-a_i)\prod_j(x-b_j)(x-\bar b_j),
$$
where $a_i\in\mathbb R$, $b_j\in\mathbb C\setminus \mathbb R$. By the Lemma, it suffices to find multiples of each of $x-a_i$ and $(x-b_j)(x-\bar b_j)$ which are $N$-extremal for a sufficiently large $N$.

These multiples are respectively $x^n-a_i^n$ and $(x^n-b_j^n)(x^n-\bar b_j^n)$ for some sufficiently large $n$.

To summarise, the desired multiple that works has the form
$$
f(x)f(\zeta x)\cdots f(\zeta^{n-1}x),
$$
where $\zeta$ is some primitive $n$-th degree root of unity. It even has integer coefficients, if $f$ has integer coefficients.