It is stated in Exercise 3.43 in Amritanshu Prasad's notes that
For two characters $\omega$ and $\eta$ of $\mathbb{F}_{q^2}^{*}$, their restrictions to $\mathbb{F}_{q}^{*}$ are equal if and only if either $\omega=\eta$ or $\omega=\eta\circ F$ where $F$ is the Frobenius automorphism of $\mathbb{F}_{q^2}^{*}$.
From my persepective, $\mathbb{F}_{q^2}^{*}$ is a cyclic group with order $q^2-1$ and a generator $c$ of $\mathbb{F}_{q^2}^{*}$ can be chosen, while $\mathbb{F}_{q}^{*}$ can be written as the cyclic group generated by $c^{q+1}$. Hence, it appears to me that $\omega|_{\mathbb{F}_{q}^{*}}=\eta|_{\mathbb{F}_{q}^{*}}$ is equivalent to $\omega(c^{q+1})=\eta(c^{q+1})$, which seems not to necessarily deduce that either $\omega(c)=\eta(c)$ or $\omega(c)=\eta(c^q)$.