Let $p$ be some prime, and consider the multiplicative group $(\mathbb{Z}_p^*, \cdot)$ and let $g$ be some generator of the group. Likewise, consider the additive group $(\mathbb{Z}_p, +)$.
The characters of additive group group $\chi_\alpha \colon \mathbb{Z}_p \to \mathbb{C}$ are defined as $\chi_\alpha(x) = \omega^{\alpha x}$, where $\omega = e^{2\pi i/p}$.
However, the characters of the multiplicative can also be seen exactly as $\{\chi_\alpha\}_{\alpha \neq 0}$ as follows: Let $\psi_{g^\alpha} \colon \mathbb{Z}_p^* \to \mathbb{C}$ be some character, and notice that if we set $\psi_{g^\alpha}(g^x) =\chi_\alpha(x)$ then we can see easily that: $$ \psi_{g^\alpha}(g^x \cdot g^y) = \chi_\alpha(x + y) = \chi_\alpha(x) \chi_\alpha(y) = \psi_{g^\alpha}(g^x) \psi_{g^\alpha}(g^y) $$ which proves that $\{\psi_{g^\alpha}\}_{\alpha \neq 0}$ are indeed homomorphic over $\mathbb{Z}_p^*$. Their orthogonality follows from $\chi_{\alpha}$ orthogonaility, and counting argument finally shows as that we found all the characters of the multiplicative group.
Now that's seems weird as it implies that the fourier coefficients of functions over the multiplicative group are just the same as over the additive group.
What's wrong here with my view?
Ok, I think I found the bug: $\mathbb{Z}^*_p$ has $p−1$ elements, whereas $\mathbb{Z}_p$ has $p$ elements. Thus $p$-roots of unity are powers of $e^{2\pi i/(p−1)}$, in contrast to powers of $e^{2\pi i/p}$ in case of the later. Thus the equivalence is true, but rather with respect to different roots...