If $P \in \mathbb {C}[X], P = X^n + a_1X_{n-1} + .. + a_{n-1}X + (-1)^n$ such that P has all the roots of same modulus, then $P(-1) \in \mathbb{R}$
It's obvious from Vieta that the modulus of all roots is $1$ also the product of all roots is $1$.
All I have to prove is $P(-1)=\overline{P(-1)}$ but I can't get it.
Let $\;\;P(z) = \prod_{k=1}^{n} (z-z_k)\;\;$ where $z_k$ are the (complex) roots. As noted in the OP $\;\;\prod_{k=1}^{n} z_k = 1\;\;$ by Vieta and, given the roots all have the same modulus, it follows that $|z_k| = 1$.
Consider now:
$$ \begin{align} \overline{P(\frac{1}{z})} & = \overline{\prod_{k=1}^{n} (\frac{1}{z}-z_k)} \\ & = \prod_{k=1}^{n} (\frac{1}{\bar z}-\bar z_k) \\ & = \prod_{k=1}^{n} (\frac{1}{\bar z}-\frac{1}{z_k}) & \text{since } z_k \bar z_k = 1 \\ & = \frac{1}{\bar z^n} \frac{1}{\prod_{k=1}^{n} z_k} (-1)^n \prod_{k=1}^{n} (\bar z - z_k)\\ & = \frac{(-1)^n}{\bar z^n} P(\bar z) & \text{since } \prod_{k=1}^{n} z_k = 1 \end{align} $$
Writing the above for $z = -1$ gives $\;\;\overline{P(-1)} = P(-1)\;$ i.e. $\;P(-1) \in \mathbb{R}$.