As a subgroup of $L^\times$, $H$ is cyclic. Hence $\gamma^k=1$ iff $k$ is a multiple of $m$. Thus $f(\gamma)=0$ implies $m|n$.
For $\Phi\in G$, we must have $f(\Phi(\gamma))=\Phi(f(\gamma))=0$, hence $\Phi(\gamma)\in H$ and there exists $k\le m$ with $\Phi(\gamma)=\gamma^k$. Since $\Phi$ has some inverse $\Psi$, gow which by the same argument $\Psi(\gamma)=\gamma^r$ for some $r$, we have $\gamma=\Psi(\Phi(\gamma))=\gamma^{rk}$ and conclude that $rk\equiv 1\pmod m$, hence $\gcd(k,m)=1$ and we can view $k$ as an element of $(\mathbb Z/m \mathbb Z)^\times$. Since exponents multiply when composing such exponentiation maps, we see that $\Phi\mapsto k$ is in fact a homomorphism from $G$ to the multiplicative group $(\mathbb Z/m \mathbb Z)^\times$. Since $L=K[\gamma]$, a $\Phi\in G$ with $\Phi(\gamma)=\gamma$ is the identity, hence the homomorphism is injective.