Is the Inner Product on an Exterior Algebra Invariant under Rotors? Take two k-vectors $x = \bigwedge_{i = 1}^k \vec{x}_i$, $y = \bigwedge_{i = 1}^k \vec{y}_i$ in $\Lambda^k(\mathbb{R}^n)$. Define the inner product between them as:
$$ \langle x, y \rangle = \det(\langle \vec{x}_i, \vec{y}_j\rangle) \equiv \star(x\wedge\star y). $$
Now we let both x and y transform under a rotor $x \mapsto RxR^{-1}$, $y \mapsto RyR^{-1}$. Does $\langle x, y \rangle$ stay constant?
Half Attempted proof:
$$ \langle RxR^{-1}, RyR^{-1} \rangle $$
$$ \overset{?}{=} \det(\langle  A \vec{x}_i, A\vec{y}_j\rangle)$$
where $A$ is a unitary rotation matrix representing the rotor 'distributed' over the constituent vectors. Not 100% sure if this is valid or possible. If it is then
$$ \langle A \vec{x}_i, A \vec{y}_j \rangle = \langle \vec{x}_i, \vec{y}_j\rangle$$
meaning each component of the Gram matrix is the same and the inner product stays constant.
If the inner product does stay constant, is there a simpler proof than the one I attempted? If it doesn't stay constant, is there a proof of that?
 A: $
\newcommand\form[1]{\langle#1\rangle}
\newcommand\rev\widetilde
$
The inner product between multivectors $X, Y$ is given in geometric algebra by the scalar product:
$$
  \form{X, Y} = \form{\rev XY}_0,
\tag{$*$}
$$
where $\rev X$ is the reversal of $X$ and $\form{\cdot}_0$ is the scalar projection. The scalar product is symmetric, from which it follows that
$$
  \form{\rev{(RX\rev R)}RY\rev R}_0
  = \form{R\rev X\rev RRY\rev R}_0
  = \form{R\rev XY\rev R}_0
  = \form{\rev XY\rev RR}_0
  = \form{\rev XY}_0,
$$
where we've used the fact that $R^{-1} = \rev R$.
Alternatively, we could also say
$$
  \form{R\rev XY\rev R}_0 = R\form{\rev XY}_0\rev R = \form{\rev XY}_0R\rev R = \form{\rev XY}_0
$$
since $x \mapsto Rx\rev R$ is grade-preserving and $\form{\cdot}_0$ is a scalar.

Assuming by $\star$ you mean the Hodge star defined by
$$
  x\wedge(\star y) = \form{x, y}I
$$
where $I$ is some chosen unit pseudoscalar, your equation $\form{x, y} = \star(x\wedge\star y)$ is incorrect for arbitrary metrics (though does work for some metric such as Euclidean metrics) and should instead be
$$
  \form{x, y} = \star^{-1}(x\wedge\star y),
$$
where $\star^{-1}$ is the inverse of the Hodge star.
I don't know if there is a straightforward way to prove ($*$) that isn't tedious or overtly abstract, but if you're willing to accept that
$$
  \star X = \rev XI,\quad \star^{-1}X = \rev{XI^{-1}},
$$
then we see
$$
  \form{x, y} = \star^{-1}(x\wedge\star y)
  = (x\wedge(\rev yI)I^{-1})^{\rev{\hphantom X}}
  = \rev{I^{-1}} (\rev Iy)\wedge\rev x
  = \rev{I^{-1}}\form{\rev Iy\rev x}_n
  = \rev{I^{-1}}\rev I\form{y\rev x}_0
  = \form{\rev xy}_0
$$
where $\form{\cdot}_n$ is pseudoscalar projection.
