Let $G$ be a finite group and $\rho \colon G \to \mathrm{GL}(n, \overline{\mathbb Q})$ be a representation. A theorem of Frobenius says that $\rho(G)$ is conjugate (in $\mathrm{GL}(n, \overline{\mathbb Q})$) to a subgroup of $\mathrm{GL}(n, \mathbb Q(\zeta_e))$, where $e$ is the exponent of $G$ and $\zeta_e$ is a primitive $e$th root of unity. We say that $\rho$ is defined over $\mathbb Q(\zeta_e)$.
Considering a cyclic group of prime order shows that Frobenius' field is in general optimal. Sometimes however we are lucky and can do better than this: consider for instance the dihedral group $D_4$ of order $8$. Its unique irreducible representation of degree $2$ maps the rotation $r$ to $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and the symmetry $s$ to $\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$, i.e., this representation is real.
We can decide whether an irreducible representation can be defined over $\mathbb R$ by calculating the Frobenius-Schur indicator of its character: if the Frobenius-Schur indicator is $1$, then the given irreducible representation can be defined over the real numbers and if it's different from $1$, then the representation is not real. If we calculate the Frobenius-Schur indicator for the irreducible representation of the quaternionic group $Q_8 = \langle a,b \ | \ a^4 = 1, \ a^2 = b^2, \ ab = b^{-1}a\rangle$ given by
$$a \mapsto \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, \qquad b \mapsto \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix},$$
we end up with $-1$, showing that this representation cannot be defined over $\mathbb R$ (in fact, it is quaternionic, which is not surprising, since we have a quaternion group). Frobenius' theorem says that we can define it over $\mathbb Q(i)$, which I already did by giving the matrices.
The following has been bothering me for a while: we can define the above representation of $Q_8$ over $\mathbb Q(\omega)$ for a primitive third root $\omega$! In fact, it is equivalent to the representation given by
$$a \mapsto \begin{pmatrix} 1+2\omega & -1 \\ -2 & -1-2\omega \end{pmatrix}, \qquad b \mapsto \begin{pmatrix}-1 & \omega^2 \\ -2\omega & 1\end{pmatrix}.$$
This seems odd to me. My question is therefore: Given a representation of a finite group and an integer $d \geq 3$, how can we decide whether it is defined over $\mathbb Q(\zeta_d)$?
For instance: is the above representation of $Q_8$ definable over $\mathbb Q(\zeta_5)$?
There is of course the obvious necessary condition that the character must take values in $\mathbb Z[\zeta_d]$. Also, as explained above, taking $d$ as the exponent of the group works.
I appreciate any help!