We let $\mathbb{K}$ be a field with characteristic $0$, $G$ a finite and abelian group, $H=\text{Aut}(G)$ and $\theta:H\rightarrow \mathbb{K}^{\times}$ a homomorphism of groups. Moreover, we denote by $T_h:\mathbb{K}[G]\rightarrow \mathbb{K}[G]$ the only linear transformation extension of $h\in H$ for every $h\in H$ (that means $T_h$ acts on $G$ the same as $h$ does, and it's defined over all of $\mathbb{K}[G]=\{\sum_{g\in G}a_gg|a_g\in\mathbb{K}\}$ as a linear transformation).
We let $g\in G$ be an element s.t for every $h\in H$, if $h(g)=g$ then $\theta(h)=1$. I have to prove that for every $h'\in H$, $v=\sum_{h\in H}\theta(h^{-1})h(g)$ is an eigenvector of $T_{h'}$, with eigenvalue $\theta(h')$.
I was able to prove that $T_{h'}(v)=\theta(h')v$, but why is v necessarily not $0$?
If it helps, we did discuss fourier transform for finite abelian groups. Thanks in advance.