Basic setting:
Let $V$ be a $k$-vector space of finite dimension and $V^*$ its dual space. Let $\bigwedge^n V$ denote the $n$-th exterior power of $V$.
Now the canonical pairing $$V \times V^{*} \to k, (v,v^*) \mapsto v^*(v)$$ gives rise to a pairing $P(,)$ on the exterior powers: $$\bigwedge\nolimits^n V \times \bigwedge\nolimits^n V^{*} \to k$$ $$(v_1\wedge\dots\wedge v_n,v^*_1 \wedge \dots \wedge v^*_n) \mapsto \det (v_i^*(v_j))_{i,j}$$ This pairing is perfect (as one can compute on a basis).
Let $G$ be a finite group whose order is not divided by the characteristic of $k$. Let $V$ now be a $G$-set, ie. as a module over the group ring $k[G]$. (or equivalently as a representation of $G$ over the field $k$).
Question: Is the induced isomorphism $$\bigwedge\nolimits^n V^* \to (\bigwedge\nolimits^n V)^*,$$ $$w^* \to (w \mapsto P(w,w^*))$$ an isomorphism of $G$-sets?
Attempt: For $w = w_1 \wedge \dots \wedge w_n \in \bigwedge\nolimits^n V$ and $w^* = w^*_1 \wedge \dots \wedge w^*_n$ I calculated:
\begin{align*} P(w,g \cdot w^*) & = \det ((g \cdot w_i^*)(w_j))_{i,j} \\ & = \det (w_i^*(g^{-1}w_j)_{i,j} \end{align*} by the definition of the dual representation. But now I can't see why this is equal to $$g \cdot P(w,w^*)$$
Is this the right attempt to show that the above isomorphism respects the $G$-action? How to proceed? Thank you for your help :-)
Thanks to Sanchez I got the answer:
First of all, we prove that the pairing on the exterior power of $V$ is invariant under the $G$-action:
\begin{align*} P(gv,gv^*) & = P(gv_1 \wedge \dots \wedge gv_n, gv_1^* \wedge \dots \wedge gv_n^*) \\ & = \det ((g \cdot v_i^*)(g \cdot v_j))_{i,j} \\ & = \det (v_i^*(g^{-1} g \cdot v_j))_{i,j} \\ & = \det (v_i^*(v_j))_{i,j} \\ & = P(v,v^*) \end{align*}
Since the pairing is an invariant $G$-form and perfect, the induced homomorphism is an isomorphism and $G$-equivariant. Let's name it $\psi: \bigwedge\nolimits^n V^* \to (\bigwedge\nolimits^n V)^*, w^* \mapsto P(-,w^*)$. Then
\begin{align*} \psi ( g \cdot w^*)(w) & = P(w,g \cdot w^*) \\ & = P(g^{-1} \cdot w, g^{-1} g \cdot w^*) \\ & = P(g^{-1} \cdot w, w^*) \\ & = g \cdot \psi(w^*)(w) \end{align*} where the last equality stems from the definition of the action of $G$ on the dual space $(\bigwedge\nolimits^n V)^*$.