Let $L/k$ be a finite separable extension and let $G$ be a finite group of automorphisms. We also fix a separable closure $k_s$ of $k$.
I want to prove that $\hom_k(L^G,k_s)=\hom_k(L,k_s)/G$, where two elements $\varphi,\psi\in\hom_k(L,k_s)$ are identified if $\varphi=\psi\circ g$ for some $g\in G$. In other words, I want to prove that $\varphi$ and $\psi$ agree on $L^G$ if and only if $\varphi=\psi\circ g$ for some $g\in G$.
This seems closely related to Galois theory but the only result I know in that could be useful is that $L/L^G$ is Galois and its Galois group is $G$ but I don't see how to use it.
Hint. For "only if". If $\phi$ and $\psi$ agree on $L^G$ then, considering them as $k$-automorphisms of $L$, $\varphi\circ\psi^{-1}$ is in the Galois group of $L|L^G$.
For "if". Check directly if $\psi$ and $\psi \circ g$ agree on $L^G$.