As $p$ is odd, the group $G=\mathbb{Z}^*_{p^n}$ is cyclic of order $p^{n-1}(p-1)$. So if $g$ is a generator of $G$, and $\omega$ is any complex number that satisfies the equation $\omega^{p^{n-1}(p-1)}=1,$
then we get a homomorphism $\chi=\chi_\omega$ from $G$ to $\mathbb{C}^*$ from the recipe
$$
\chi_\omega(g^k)=\omega^k
$$
for all $k=0,1,2,\ldots,p^{n-1}(p-1)-1$.
There are $p^{n-1}(p-1)$ choices for $\omega$, and each choice gives you a different character. It is not too hard to see that there are no others.
If you are only interested in characters that take values $\pm1$ only, then you must pick $\omega=\pm1$. The choice $\omega=+1$ gives you the trivial character (everything is mapped to one), and the choice $\omega=-1$ gives the character $\eta=\chi_{-1}$ that you described:
$$
\eta(g^k)=(-1)^k=\begin{cases}+1,\ \text{if $k$ is even,}\\-1,\ \text{if $k$ is odd}.\end{cases}
$$
So your $\eta$ is a multiplicative character all right (the square are exactly the even powers of $g$, as the order of $g$, $p^{n-1}(p-1)$, is an even number.
For example, when $p^n=3^2=9$, then $|G|=3^{2-1}(3-1)=6$, and it is easy to see that we can use $g=2$. If we select $\omega$ to be a primitive sixth root of unity, say $\omega=e^{\pi i/3}=(\sqrt3+i)/2$, then we get the character $\chi=\chi_\omega$:
$$
\begin{eqnarray*}
\chi(1)=\chi(2^0)&=\omega^0=1,\\
\chi(2)=\chi(2^1)&=\omega^1=\omega,\\
\chi(4)=\chi(2^2)&=\omega^2,\\
\chi(8)=\chi(2^3)&=\omega^3=-1,\\
\chi(7)=\chi(2^4)&=\omega^4=-\omega,\\
\chi(5)=\chi(2^5)&=\omega^5=\overline{\omega}.
\end{eqnarray*}
$$
This time $\chi(0),\chi(3)$ and $\chi(6)$ are left undefined as those numbers are divisible by three. Sometimes a convention is used, when those are defined to be zero. Those three residue classes do not belong to the group $G$, but number theorists sometimes want to define these functions outside of $G$ as well.
This time you get the quadratic character $\eta$ as the third power $\chi^3$.
Hope this helps. Ask for clarifications if/where needed.
Yes this holds for Fermat primes. You need the following observation. This is sometimes called the $N/C$ Theorem.
Lemma Let $G$ be a group with a subgroup $H$, then $N_G(H)/C_G(H)$ embeds homomorphically into Aut$(H)$.
Here $N_G(H)=\{g \in G : gH=Hg\}$, the normalizer of $H$ in $G$ and $C_G(H)=\{g \in G : gh=hg \text { for all } h\in H\}$, the centralizer of $H$ in $G$.
Now let us apply this to the situation where $N=H$ is normal, $|N|=$ Fermat-prime and $|G|$ is odd. Then $N_G(N)=G$ because of the normality of $N$. And since $N$ is cyclic, |Aut$(N)|$ is a power of $2$ (in fact $|N|-1$). It follows from the $N/C$ theorem that $|G/C_G(N)|$ is a power of $2$. But obviously it also divides $|G|$, which is odd. This can only be when $G=C_G(N)$, that is $N \subseteq Z(G)$.
So how do you prove the lemma? Let me give a sketch and leave the details with you: $N_G(H)$ acts as automorphisms on $H$ by conjugation. The kernel of the action is $C_G(H)$.
Generalizations:
Proposition 1 Let $G$ be a finite group, $N$ a normal subgroup of prime order $p$, with gcd$(|G|,p-1)=1$, then $N \subseteq Z(G)$.
Note that this holds for $p=2$: a normal subgroup of order $2$ must be central.
Proposition 2 Let $G$ be a finite group, $N$ a normal subgroup of prime order $p$, where $p$ is the smallest prime dividing $|G|$. Then $N \subseteq Z(G)$.
Best Answer
If $\chi:G\to\Bbb{C}^*$, $G$ a finite abelian group, is a character of order two, it follows that $1=\chi(g)^2=\chi(g^2)$ for all $g\in G$. This implies that
If we furthermore know that $G$ is cyclic of an even order $n=2m$, then
So if $\chi$ has order two, then
As Hempelicious pointed out, the character group of an abelian group is isomorphic to the group itself. But, the isomorphism is not natural. In other words, it relies on our ability to write a finite abelian group as a direct product of cyclic groups. Such a decomposition is not unique, and this leads to complications similar to the well known result that the dual of a f.d. vector space is isomorphic to the vector space itself, but to write down the isomorphism you need to specify a basis. In sharp contrast to the simpler fact that the double dual is naturally isomorphic to the vector space we started with.