Lifting multiplicative characters over finite fields Let $\mathbb{F}_{q^n}$ be a finite field, where $q$ is a power of a prime.
If $\eta$ is a multiplicative character of $\mathbb{F}_{q}$, then one can lift $\eta$ to $\mathbb{F}_{q^n}$ defining $\overline{\eta}(w) = \eta(N(w))$, where $N$ is the norm from $\mathbb{F}_{q^n}$ to $\mathbb{F}_{q}$. You can also restrict $\overline{\eta}$ to $\mathbb{F}_{q}$ creating a new character over $\mathbb{F}_{q}$, $\eta^\ast$. It is a fact that it need not to be $\eta = \eta^\ast$. Indeed, if $\eta$ has order $e$ (a divisor of $q-1$), then $\eta^\ast$ has order $\dfrac{e}{gcd(e,n)}$. In particular $\eta = \eta^\ast$ if and only if the order of $\eta$ divides $n$. One then finds that there are $gcd(n,q-1)$ characters of $\mathbb{F}_{q}$ such that $\eta^\ast$ is trivial. 
My question is, how many multiplicative characters $\eta$ are there such that $\overline{\eta}$ and $\eta^\ast$ are trivial? For me it transpires that there is just one, the trivial character, however I'm not sure and I can't give a formal proof of this.
 A: *

*Let $\phi \in \text{Aut}(\mathbf{F}_{q^n}), \phi(\alpha) = \alpha^q$ the Frobenius automorphism and $\langle \phi \rangle$ the cyclic group it generates. As $\mathbf{F}_{q}$ is the splitting field of $X^{q}-X$ then $\mathbf{F}_{q} = \{ \alpha \in \mathbf{F}_{q^n}, \phi(a) = a\}$ is the fixed field of $\langle \phi \rangle$ (so that $\mathbf{F}_{q^n}/\mathbf{F}_{q}$ is a Galois extension, and for $\alpha \in \mathbf{F}_{q^n}$, its minimal polynomial over $\mathbf{F}_{q}$ is
$f(X) = \prod_{\beta \in \langle \phi \rangle \alpha}(X-\beta)= \prod_{m=1}^d (X-\alpha^{q^m}) \in \mathbf{F}_q[X]$)


*Thus the field norm is $$N(\alpha)= N_{\mathbf{F}_{q^n}/\mathbf{F}_{q}}(\alpha) = \prod_{m=1}^n \phi^m (\alpha) = \prod_{m=1}^n \alpha^{q^m} = \alpha^{(q^n-1)/(q-1)}$$
And for $a \in \mathbf{F}_{q}$, $\eta^*(a) = \eta(N(a))= \eta(a^n) = \eta(a)^n$ so that $\eta = \eta^*$ iff $n \equiv 1 \bmod \text{ord}(\eta)$.


*As $X^{q^n-1}-1$ has $q^n-1$ different roots in $\mathbf{F}_{q^n}$ then so does $X^{(q^n-1)(q-1)}-1$. Thus, since $N$ is an homormophim $\mathbf{F}_{q^n}^\times\to \mathbf{F}_{q}^\times$ we find
$$\# \ker(N) = \frac{q^n-1}{q-1}, \qquad \# \text{Im}(N) = \frac{\# \mathbf{F}_{q^n}^\times}{\# \ker(N)} = q-1 \quad \implies \quad N \text{ is surjective}$$
which means $\overline{\eta}$ is the trivial character iff $\eta$ is trivial.
