Which metrics on exterior power are induced from metrics on the base? $\newcommand{\id}{\text{id}}$
$\newcommand{\Hom}{\text{Hom}}$
Let $V$ be a $d$-dimensional real vector space, and let $2 \le k \le d-1$. Every inner product on $V$ induces an inner product on $\Lambda^k V$, in the following way
$$ \langle v_1 \wedge \dots \wedge v_k , w_1 \wedge \dots \wedge w_k \rangle:=\det (\langle v_i ,w_j \rangle). \tag{1}$$
Question:
What are necessary and sufficient conditions on an inner product on $\Lambda^k V$ to to be induced from a product on $V$ as in $(1)$?
(I restricted $k \neq 1,d$ because then of course, any metric is induced by a metric on the base).
I think there should be some "compatibility" or "symmetry" conditions, but I am not sure how to formulate them. 

Edit:
If there exist an inducing product at the base, this product is unique. Perhaps we can construct an "inverse map" which is defined on all products on $\Lambda^k V$, and see when the result is an honest inner product on $V$.
Proof of uniqueness:
Suppose $g_1,g_2$ are inner products on $V$, which induce the same product on $\Lambda^k V$. 
Let $A:(V,g_1) \to (V,g_2)$ be an isometry. Then the induced exterior map $$\bigwedge ^k A:(\Lambda^k V,\Lambda^k g_1) \to (\Lambda^k V,\Lambda^k g_2)$$  
is an isometry, where $\Lambda^k g_i$ is the metric on $ \Lambda^k V$ induced by $g_i$ via $(1)$. Since by assumption $\Lambda^k g_1=\Lambda^k g_2$, we get that
$$\bigwedge ^k A:(\Lambda^k V,\Lambda^k g_1) \to (\Lambda^k V,\Lambda^k g_1)$$  
is an isometry, i.e.
$$ \id_{\Lambda_k(V)}=(\bigwedge ^k A)^T \circ \bigwedge ^k A= \bigwedge ^k A^T \circ \bigwedge ^k A=\bigwedge ^k A^T A,$$
where the transpose in $A^T$ is taken with respect to the metric $g_1$. Denote $S=A^TA$. Then $S \in \Hom(V,V)$ is symmetric (w.r.t. $g_1$) and positive-definite, and satisfies $  \id_{\Lambda_k(V)}  =\bigwedge^k S $.
This implies $S=\id_V$. (For a short proof, see this answer).
So, we have obtained $A^TA=\id_V$; Remembering the transpose was taken w.r.t $g_1$, this shows $A$ is an automorphism of $(V,g_1)$. Recalling that we assumed $A:(V,g_1) \to (V,g_2)$ was an isometry, this shows $g_1=g_2$.
 A: This is a partial answer, for the case $k=d-1$:
In this case, $\Lambda^{d-1} V$ and $V$ have the same dimension, so it is reasonable to expect any metric on $\Lambda^{d-1} V$ comes from a metric on $V$. We shall prove this is indeed the case.
First, we reformulate the problem:
A choice of a metric on $V$ is equivalent to a choise of a linear isomorphism $ g:V \to V^*$ that satisfies
$$ g(v)(w)=g(w)(v) \, \, \text{and}\, \,g(v)(v) \ge 0 \, \, \text{with equality only when } \, v=0. \tag{1}$$
The equivalence is via $g(v)(w):= \langle v,w \rangle$. Using this perspective, the induced metric on $\Lambda^{k} V$ induced by $g$ is $\Lambda^kg:\Lambda^{k} V \to \Lambda^{k} (V^*) \cong  (\Lambda^{k} V)^*$.
So, the question becomes the following:

For which maps $h:\Lambda^{k} V \to  (\Lambda^{k} V)^*$ which are symmetric and positive in the sense of $(1)$, there exist a symmetric and positive $g$ such that $h=\Lambda^kg$?

In the case of $k=d-1$, the answer is that all such $h$ come from $g$'s at the base.
Indeed, it follows from this question that every orientation-preserving invertible linear map $\bigwedge^{d-1} V \to \bigwedge^{d-1} V^*$ equals $\Lambda^{d-1}A$ for some invertible $A:V \to V^*$. 
A metric (regarded as a map $V \to V^*$) is always orientation-preserving (it maps an orthonormal basis $e_i$ to its dual basis), so our $h$ equals  $\Lambda^{d-1}g$ for some $g$. We now need to prove $g$ is symmetric and positive.
The symmetry is not hard to prove. Also, it is quite easy to see $g$ must be definite (negative or positive). If $d-1$ is odd, it must be positive-definite. When $d-1$ is even $\Lambda^{d-1}g=\Lambda^{d-1}(-g)$, so we can always choose the "positive root".
This finishes the proof.
