Isomorphic representations on exterior powers Exercise from F+H, Exercise 1.3:

Let $\rho : G \rightarrow GL(V)$ be any representation of the finite group $G$ on a $n$-dimensional vector space $V$ and suppose that for any $g \in G$ the determinant of $\rho(g)$ is 1. Show that $\bigwedge^k V$ and $\bigwedge^{n-k} V^*$ are isomorphic as representations of G

For the life of me I can't figure out. I know that $\bigwedge^k V$ and $\bigwedge^{n-k} V^*$ are isomorphic as spaces, but why are they isomorphic as representations, I have no idea. I suspect it has something to do with the determinant being 1, but...
 A: For any field $F$ and vector space $V$ over $F$ there is a map
$$\wedge^k V \rightarrow \mathrm{Hom}_F(\wedge^{n-k} V, \wedge^n V)$$ defined by sending an element $x \in \wedge^k V$ to the linear map $y \mapsto x \wedge y$ which is wedge product with $x$. If $V$ is $n$-dimensional then the determinant induces an isomorphism of vector spaces
$$\wedge^n V \rightarrow F$$ so that by composing we obtain a map of vector spaces
$$\phi:\wedge^k V \rightarrow \mathrm{Hom}_F(\wedge^{n-k} V, \wedge^n V)\rightarrow \mathrm{Hom}_F(\wedge^{n-k} V, F)$$ that you can show is actually an isomorphism. For any $g \in GL(V)$ we have $\phi(gx)=\mathrm{det}(g) g \phi(x)$, so if $\mathrm{det}(g)=1$ this reads $\phi(gx)=g\phi(x)$, as desired.
A: Consider the map $\phi:\bigwedge^kV\to \bigwedge^{n-k}V^*$ that takes $x\in \bigwedge^k V$ to the map $y\mapsto x\wedge y$ for $y\in \bigwedge^{n-k}V$. Since the two vector spaces have the same dimension, to see that it is a vector space isomorphism it suffices to check that $\ker \phi=\mathbf{0}$. This is indeed the case, as you can check ($x\wedge y=0$ for all $y$ implies $x=0$).
So it only remains to check that it is a representation map i.e. that
$$g\bigr(\phi(x)\bigr)=\phi\bigr(gx\bigr).$$
To see this, let $y\in \bigwedge^{n-k}V$. We have $$\phi(gx)(y)=gx\wedge y=gx\wedge gg^{-1}y=\det(\rho(g))\, x\wedge g^{-1}y=x\wedge {g^{-1}y},$$ this is indeed $g(\phi (x))(y)$ (Check out the dual representation).
