Invariant characterization of isometric embeddings $\newcommand{\Cof}{\operatorname{Cof}}$
$\newcommand{\Det}{\operatorname{Det}}$
$\newcommand{\Lam}{\operatorname{\Lambda}}$
Motivation (and the "classic" case):
I am trying to find a coordinate-free criterion to determine when a linear map is an isometric embedding of an inner product space in a space of higher dimension. 
Such a criterion exists when the dimensions are equal. Let $V,W$ be oriented inner product spaces, of the same dimension $d$. 
Let $\,A: V \to W$ be a linear map. Then one can define invariantly the cofactor and the determinant of $A$ as follows: $$(1) \Det A :=  \star_W^d \circ \bigwedge^d A \circ \star_V^0 ,$$ $$(2) \Cof A = (-1)^{d-1} \star_W^{d-1} \circ \bigwedge^{d-1} A \circ \star_V^1,$$ 
Note that $\Det A \in \mathbb{R}$,$\, \Cof A:V \to W$, (I keep the degree explicit in the notation of the Hodge-dual, i.e $\star_V^p:\Lam_p(V) \to \Lam_{d-p}(V)$). 
The well-known matrix identity $$\Cof X \cdot X^T = \det X \cdot \operatorname{Id}$$ can be invariantly written as $$(*) \,  \Cof A \circ  A^T= \Det A \cdot \operatorname{Id}_W$$
Immediate observation: $(*)$ implies that $A$ is an isometry if and only if $$(3) \Det A \neq 0 \text{ and} \, \Det A \cdot A= \Cof A$$
My question is about the validity of a possible approach of generalizing criterion $(3)$ to the case where $\dim V < \dim W$. 

The case $\dim V < \dim W$:
Stage I: generalizing identity $(*)$.
Assume $\dim V=d < n=\dim W$. Definitions $(1),(2)$  for the cofactor and the determinant now gives objects of different types:
(a) $\Cof A:V \to \bigwedge^{n-d+1}W$, so $\Cof A \circ  A^T:W \to \bigwedge^{n-d+1}W$.
(b) $\Det A \in \bigwedge^{n-d}W$.
The natural generalization of identity $(*)$ should therefore be:
$$(**) \Cof A \circ  A^T(w)= \Det A \wedge w \, ,\,\forall w \in W.$$ 
(This is an equality of elements in $\bigwedge^{n-d+1}W $).
Stage II: generalizing condition $(3)$.
I think $A$ will be an isometric embedding if and only if $$ (3)' \Cof A \, \text{is injective and} \, \Det A \wedge A(v) = \Cof A(v) \, \, \forall v \in V  $$ (This is an equality of elements in $\bigwedge^{n-d+1}W $).
Indeed, putting $w=Av$ in $(**)$, one gets that for any linear map $A$, $$ \Cof A \circ  A^T(Av)= \Det A \wedge Av, $$ so $A$ satisfies condition $(3)'$ if and only if $$ \Cof A \, \text{is injective and} \, \Cof A \circ  (A^TA) = \Cof A$$ which holds iff $A^TA=\operatorname{Id}_V$ 
(which is equivalent to $A$ being an isometry).
The remaining questions (which are needed for completion of this endeavour) are:
(1)  Does identity $(**)$ hold? (It probably holds "up to sign", i.e both $d,n$ should enter the new definition of the cofactor in a manner which should be determined  )
(2) In both cases, part of the conditions were "non-triviality" , in $(3)$ it was $\det A \neq 0$ which is equivalent to the injectivity of $A$. In $(3)'$ it was injectivity of $\Cof A:V \to \bigwedge^{n-d+1}W$. Is it true that injectivity of $A$ implies injectivity of $\Cof A$? (and what about the converse?)
 A: Here is a proof of $\left(  \ast\ast\right)  $. As you correctly admitted,
part of the problem is to define the signs that enter into
$\operatorname*{Cof}A$. Let me actually start from scratch and introduce all
notations that will be needed:
For any $m\in\mathbb{N}$, we shall write $\left[  m\right]  $ for the set
$\left\{  1,2,\ldots,m\right\}  $.
Let $d$ and $n$ be two nonnegative integers. Let $A\in\mathbf{k}^{n\times d}$
be an $n\times d$-matrix over a commutative ring $\mathbf{k}$. Regard $A$ as a
$\mathbf{k}$-linear map $\mathbf{k}^{d}\rightarrow\mathbf{k}^{n}$.
Let $\left(  f_{1},f_{2},\ldots,f_{d}\right)  $ be the standard basis of the
$\mathbf{k}$-module $\mathbf{k}^{d}$. (Thus, $f_{i}$ is the vector whose
$i$-th entry is $1$ and whose all other entries are $0$.)
Let $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ be the standard basis of the
$\mathbf{k}$-module $\mathbf{k}^{n}$. For every subset $I=\left\{  i_{1}
<i_{2}<\cdots<i_{k}\right\}  $ of $\left[  n\right]  $, we let $e_{I}$ be the
vector $e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k}}\in\wedge
^{k}\left(  \mathbf{k}^{n}\right)  $.
If $U\subseteq\left[  n\right]  $ and $V\subseteq\left[  d\right]  $, then we
let $A_{U,V}$ denote the submatrix of $A$ obtained by removing all rows except
for those in $U$ and all columns except for those in $V$. (In more rigorous
terms: We let $A_{U,V}$ be the matrix $\left(  a_{u_{i},v_{j}}\right)  _{1\leq
i\leq\left\vert U\right\vert ,\ 1\leq j\leq\left\vert V\right\vert }$, where
$A$ is written in the form $\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq
j\leq d}$, and where the sets $U$ and $V$ are written in the forms $U=\left\{
u_{1}<u_{2}<\cdots<u_{\left\vert U\right\vert }\right\}  $ and $V=\left\{
v_{1}<v_{2}<\cdots<v_{\left\vert V\right\vert }\right\}  $.)
If $S$ is a finite set of integers, then $\sum S$ shall mean the sum of the
elements of $S$.
Regard the $d\times n$-matrix $A^{T}\in\mathbf{k}^{d\times n}$ as a
$\mathbf{k}$-linear map $\mathbf{k}^{n}\rightarrow\mathbf{k}^{d}$.
Define an element $\operatorname*{Det}A\in\wedge^{n-d}\left(  \mathbf{k}
^{n}\right)  $ by $\operatorname*{Det}A=\sum_{\substack{I\subseteq\left[
n\right]  ;\\\left\vert I\right\vert =n-d}}\left(  -1\right)  ^{\sum
I-\left\vert I\right\vert }\det\left(  A_{\left[  n\right]  \setminus
I,\left[  d\right]  }\right)  e_{I}$.
Define a $\mathbf{k}$-linear map $\operatorname*{Cof}A:\mathbf{k}
^{d}\rightarrow\wedge^{n-d+1}\left(  \mathbf{k}^{n}\right)  $ by setting
$\left(  \operatorname*{Cof}A\right)  \left(  f_{i}\right)  =\sum
_{\substack{J\subseteq\left[  n\right]  ;\\\left\vert J\right\vert
=n-d+1}}\left(  -1\right)  ^{i+\sum J}\det\left(  A_{\left[  n\right]
\setminus J,\left[  d\right]  \setminus\left\{  i\right\}  }\right)  e_{J}$.
Theorem 1. For every $w\in\mathbf{k}^{n}$, we have $\left(  \left(
\operatorname*{Cof}A\right)  \circ A^{T}\right)  \left(  w\right)  =\left(
\operatorname*{Det}A\right)  \wedge w$ in $\wedge\left(  \mathbf{k}
^{n}\right)  $.
Proof of Theorem 1. Let $w\in\mathbf{k}^{n}$. We must prove the equality
$\left(  \left(  \operatorname*{Cof}A\right)  \circ A^{T}\right)  \left(
w\right)  =\left(  \operatorname*{Det}A\right)  \wedge w$. This equality is
$\mathbf{k}$-linear in $w$; thus, we can WLOG assume that $w$ is an element of
the basis $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ of $\mathbf{k}^{n}$. In
other words, we can WLOG assume that $w=e_{k}$ for some $k\in\left[  n\right]
$. Assume this, and consider this $k$.
Write the matrix $A$ in the form $A=\left(  a_{i,j}\right)  _{1\leq i\leq
n,\ 1\leq j\leq d}$. Thus, $A^{T}=\left(  a_{j,i}\right)  _{1\leq i\leq
d,\ 1\leq j\leq n}$.
From $w=e_{k}$, we obtain $A^{T}\left(  w\right)  =A^{T}\left(  e_{k}\right)
=\sum_{i\in\left[  d\right]  }a_{k,i}f_{i}$ (since $A^{T}=\left(
a_{j,i}\right)  _{1\leq i\leq d,\ 1\leq j\leq n}$). Now,
$\left(  \left(  \operatorname*{Cof}A\right)  \circ A^{T}\right)  \left(
w\right)  $
$=\left(  \operatorname*{Cof}A\right)  \underbrace{\left(  A^{T}\left(
w\right)  \right)  }_{=\sum_{i\in\left[  d\right]  }a_{k,i}f_{i}}=\left(
\operatorname*{Cof}A\right)  \left(  \sum_{i\in\left[  d\right]  }a_{k,i}
f_{i}\right)  $
$=\sum_{i\in\left[  d\right]  }a_{k,i}\underbrace{\left(  \operatorname*{Cof}
A\right)  \left(  f_{i}\right)  }_{\substack{=\sum_{\substack{J\subseteq
\left[  n\right]  ;\\\left\vert J\right\vert =n-d+1}}\left(  -1\right)
^{i+\sum J}\det\left(  A_{\left[  n\right]  \setminus J,\left[  d\right]
\setminus\left\{  i\right\}  }\right)  e_{J}\\\text{(by the definition of
}\operatorname*{Cof}A\text{)}}}$
$=\sum_{i\in\left[  d\right]  }a_{k,i}\sum_{\substack{J\subseteq\left[
n\right]  ;\\\left\vert J\right\vert =n-d+1}}\left(  -1\right)  ^{i+\sum
J}\det\left(  A_{\left[  n\right]  \setminus J,\left[  d\right]
\setminus\left\{  i\right\}  }\right)  e_{J}$
$=\sum_{\substack{J\subseteq\left[  n\right]  ;\\\left\vert J\right\vert
=n-d+1}}\left(  \sum_{i\in\left[  d\right]  }\left(  -1\right)  ^{i+\sum
J}a_{k,i}\det\left(  A_{\left[  n\right]  \setminus J,\left[  d\right]
\setminus\left\{  i\right\}  }\right)  \right)  e_{J}$
$=\sum_{\substack{J\subseteq\left[  n\right]  ;\\\left\vert J\right\vert
=n-d+1}}\left(  -1\right)  ^{\sum J}\left(  \sum_{i\in\left[  d\right]
}\left(  -1\right)  ^{i}a_{k,i}\det\left(  A_{\left[  n\right]  \setminus
J,\left[  d\right]  \setminus\left\{  i\right\}  }\right)  \right)  e_{J}$
(1) $=\sum_{\substack{J\subseteq\left[  n\right]  ;\\\left\vert
J\right\vert =n-d+1;\\k\in J}}\left(  -1\right)  ^{\sum J}\left(  \sum
_{i\in\left[  d\right]  }\left(  -1\right)  ^{i}a_{k,i}\det\left(  A_{\left[
n\right]  \setminus J,\left[  d\right]  \setminus\left\{  i\right\}  }\right)
\right)  e_{J}$
$+\sum_{\substack{J\subseteq\left[  n\right]  ;\\\left\vert J\right\vert
=n-d+1;\\k\notin J}}\left(  -1\right)  ^{\sum J}\left(  \sum_{i\in\left[
d\right]  }\left(  -1\right)  ^{i}a_{k,i}\det\left(  A_{\left[  n\right]
\setminus J,\left[  d\right]  \setminus\left\{  i\right\}  }\right)  \right)
e_{J}$
(here we have split the sum into two subsums, according to whether $k\in J$ or
$k\notin J$).
If $J$ is a finite set of integers, and if $k$ is an integer, then $\#\left(
J<k\right)  $ shall denote the number of all elements of $J$ that are smaller
than $k$. Similarly, $\#\left(  J>k\right)  $ shall denote the number of all
elements of $J$ that are greater than $k$.
But every $\left(  n-d+1\right)  $-element subset $J$ of $\left[  n\right]  $
satisfying $k\in J$ satisfies
(2) $\sum_{i\in\left[  d\right]  }\left(  -1\right)  ^{i+\#\left(  \left[
n\right]  \setminus J<k\right)  +1}a_{k,i}\det\left(  A_{\left[  n\right]
\setminus J,\left[  d\right]  \setminus\left\{  i\right\}  }\right)
=\det\left(  A_{\left(  \left[  n\right]  \setminus J\right)  \cup\left\{
k\right\}  ,\left[  d\right]  }\right)  $
(by Laplace expansion, applied to the matrix $A_{\left(  \left[  n\right]
\setminus J\right)  \cup\left\{  k\right\}  ,\left[  d\right]  }$, with
respect to the $\#\left(  \left[  n\right]  \setminus J<k\right)  +1$-th row
of this matrix (which used to be the $k$-th row of $A$)). Hence, every
$\left(  n-d+1\right)  $-element subset $J$ of $\left[  n\right]  $ satisfying
$k\in J$ satisfies
(3) $\sum_{i\in\left[  d\right]  }\left(  -1\right)  ^{i}a_{k,i}
\det\left(  A_{\left[  n\right]  \setminus J,\left[  d\right]  \setminus
\left\{  i\right\}  }\right)  =\left(  -1\right)  ^{\#\left(  \left[
n\right]  \setminus J<k\right)  +1}\det\left(  A_{\left(  \left[  n\right]
\setminus J\right)  \cup\left\{  k\right\}  ,\left[  d\right]  }\right)  $
(this follows by multiplying both sides of (2) with $\left(  -1\right)
^{\#\left(  \left[  n\right]  \setminus J<k\right)  +1}$).
Furthermore, every $\left(  n-d+1\right)  $-element subset $J$ of $\left[
n\right]  $ satisfying $k\notin J$ satisfies
(4) $\sum_{i\in\left[  d\right]  }\left(  -1\right)  ^{i}a_{k,i}
\det\left(  A_{\left[  n\right]  \setminus J,\left[  d\right]  \setminus
\left\{  i\right\}  }\right)  =0$
(by Laplace expansion, applied to the matrix $A_{\left[  n\right]  \setminus
J,\left[  d\right]  }$ with an extra copy of the $k$-th row of $A$ added to
its top; the result is $0$ because the resulting matrix has this $k$-th row
appear twice inside it). Now, (1) becomes
$\left(  \left(  \operatorname*{Cof}A\right)  \circ A^{T}\right)  \left(
w\right)  $
$=\sum_{\substack{J\subseteq\left[  n\right]  ;\\\left\vert J\right\vert
=n-d+1;\\k\in J}}\left(  -1\right)  ^{\sum J}\underbrace{\left(  \sum
_{i\in\left[  d\right]  }\left(  -1\right)  ^{i}a_{k,i}\det\left(  A_{\left[
n\right]  \setminus J,\left[  d\right]  \setminus\left\{  i\right\}  }\right)
\right)  }_{\substack{=\left(  -1\right)  ^{\#\left(  \left[  n\right]
\setminus J<k\right)  +1}\det\left(  A_{\left(  \left[  n\right]  \setminus
J\right)  \cup\left\{  k\right\}  ,\left[  d\right]  }\right)  \\\text{(by
(3))}}}e_{J}$
$+\sum_{\substack{J\subseteq\left[  n\right]  ;\\\left\vert J\right\vert
=n-d+1;\\k\notin J}}\left(  -1\right)  ^{\sum J}\underbrace{\left(  \sum
_{i\in\left[  d\right]  }\left(  -1\right)  ^{i}a_{k,i}\det\left(  A_{\left[
n\right]  \setminus J,\left[  d\right]  \setminus\left\{  i\right\}  }\right)
\right)  }_{\substack{=0\\\text{(by (4))}}}e_{J}$
$=\sum_{\substack{J\subseteq\left[  n\right]  ;\\\left\vert J\right\vert
=n-d+1;\\k\in J}}\underbrace{\left(  -1\right)  ^{\sum J}\left(  -1\right)
^{\#\left(  \left[  n\right]  \setminus J<k\right)  +1}}_{=\left(  -1\right)
^{\sum J+\#\left(  \left[  n\right]  \setminus J<k\right)  +1}}\det\left(
A_{\left(  \left[  n\right]  \setminus J\right)  \cup\left\{  k\right\}
,\left[  d\right]  }\right)  e_{J}$
$=\sum_{\substack{J\subseteq\left[  n\right]  ;\\\left\vert J\right\vert
=n-d+1;\\k\in J}}\left(  -1\right)  ^{\sum J+\#\left(  \left[  n\right]
\setminus J<k\right)  +1}\det\left(  A_{\left(  \left[  n\right]  \setminus
J\right)  \cup\left\{  k\right\}  ,\left[  d\right]  }\right)  e_{J}$
$=\sum_{\substack{I\subseteq\left[  n\right]  ;\\\left\vert I\right\vert
=n-d;\\k\notin I}}\left(  -1\right)  ^{\sum\left(  I\cup\left\{  k\right\}
\right)  +\#\left(  \left[  n\right]  \setminus\left(  I\cup\left\{
k\right\}  \right)  <k\right)  +1}\underbrace{\det\left(  A_{\left(  \left[
n\right]  \setminus\left(  I\cup\left\{  k\right\}  \right)  \right)
\cup\left\{  k\right\}  ,\left[  d\right]  }\right)  }_{\substack{=\det\left(
A_{\left[  n\right]  \setminus I,\left[  d\right]  }\right)  \\\text{(since
}\left(  \left[  n\right]  \setminus\left(  I\cup\left\{  k\right\}  \right)
\right)  \cup\left\{  k\right\}  =\left[  n\right]  \setminus I\text{)}
}}\underbrace{e_{I\cup\left\{  k\right\}  }}_{\substack{=\left(  -1\right)
^{\#\left(  I>k\right)  }e_{I}\wedge e_{k}\\\text{(because the sign of the
permutation}\\\text{that takes }e_{I\cup\left\{  k\right\}  }\text{ to }
e_{I}\wedge e_{k}\text{ is }\left(  -1\right)  ^{\#\left(  I>k\right)
}\text{)}}}$
(here, we have substituted $I\cup\left\{  k\right\}  $ for $J$ in the sum,
because every $\left(  n-d+1\right)  $-element subset $J$ of $\left[
n\right]  $ satisfying $k\in J$ must have the form $I\cup\left\{  k\right\}  $
for some $\left(  n-d\right)  $-element subset $I$ of $\left[  n\right]  $
satisfying $k\notin I$)
$=\sum_{\substack{I\subseteq\left[  n\right]  ;\\\left\vert I\right\vert
=n-d;\\k\notin I}}\left(  -1\right)  ^{\sum\left(  I\cup\left\{  k\right\}
\right)  +\#\left(  \left[  n\right]  \setminus\left(  I\cup\left\{
k\right\}  \right)  <k\right)  +1}\det\left(  A_{\left[  n\right]  \setminus
I,\left[  d\right]  }\right)  \left(  -1\right)  ^{\#\left(  I>k\right)
}e_{I}\wedge e_{k}$
$=\sum_{\substack{I\subseteq\left[  n\right]  ;\\\left\vert I\right\vert
=n-d;\\k\notin I}}\underbrace{\left(  -1\right)  ^{\sum\left(  I\cup\left\{
k\right\}  \right)  +\#\left(  \left[  n\right]  \setminus\left(
I\cup\left\{  k\right\}  \right)  <k\right)  +1+\#\left(  I>k\right)  }
}_{\substack{=\left(  -1\right)  ^{\sum I-\left\vert I\right\vert
}\\\text{(since }\sum\left(  I\cup\left\{  k\right\}  \right)  +\#\left(
\left[  n\right]  \setminus\left(  I\cup\left\{  k\right\}  \right)
<k\right)  +1+\#\left(  I>k\right)  \\\equiv\underbrace{\sum\left(
I\cup\left\{  k\right\}  \right)  }_{=\sum I+k}+\#\left(  \left[  n\right]
\setminus\left(  I\cup\left\{  k\right\}  \right)  <k\right)
+1-\underbrace{\#\left(  I>k\right)  }_{\substack{=\left\vert I\right\vert
-\#\left(  I<k\right)  \\\text{(since }k\notin I\text{)}}}\\=\sum
I+k+\#\left(  \left[  n\right]  \setminus\left(  I\cup\left\{  k\right\}
\right)  <k\right)  +1-\left\vert I\right\vert +\#\left(  I<k\right)  \\=\sum
I+k+1-\left\vert I\right\vert +\underbrace{\#\left(  \left[  n\right]
\setminus\left(  I\cup\left\{  k\right\}  \right)  <k\right)  +\#\left(
I<k\right)  }_{=k-1}\\=\sum I+k+1-\left\vert I\right\vert +k-1=\sum
I+2k-\left\vert I\right\vert \\\equiv\sum I-\left\vert I\right\vert
\operatorname{mod}2\text{)}}}\det\left(  A_{\left[  n\right]  \setminus
I,\left[  d\right]  }\right)  e_{I}\wedge e_{k}$
$=\sum_{\substack{I\subseteq\left[  n\right]  ;\\\left\vert I\right\vert
=n-d;\\k\notin I}}\left(  -1\right)  ^{\sum I-\left\vert I\right\vert }
\det\left(  A_{\left[  n\right]  \setminus I,\left[  d\right]  }\right)
e_{I}\wedge e_{k}$.
Comparing this with
$\underbrace{\operatorname*{Det}A}_{=\sum_{\substack{I\subseteq\left[
n\right]  ;\\\left\vert I\right\vert =n-d}}\left(  -1\right)  ^{\sum
I-\left\vert I\right\vert }\det\left(  A_{\left[  n\right]  \setminus
I,\left[  d\right]  }\right)  e_{I}}\wedge\underbrace{w}_{=e_{k}}$
$=\left(  \sum_{\substack{I\subseteq\left[  n\right]  ;\\\left\vert
I\right\vert =n-d}}\left(  -1\right)  ^{\sum I-\left\vert I\right\vert }
\det\left(  A_{\left[  n\right]  \setminus I,\left[  d\right]  }\right)
e_{I}\right)  \wedge e_{k}$
$=\sum_{\substack{I\subseteq\left[  n\right]  ;\\\left\vert I\right\vert
=n-d}}\left(  -1\right)  ^{\sum I-\left\vert I\right\vert }\det\left(
A_{\left[  n\right]  \setminus I,\left[  d\right]  }\right)  e_{I}\wedge
e_{k}$
$=\sum_{\substack{I\subseteq\left[  n\right]  ;\\\left\vert I\right\vert
=n-d;\\k\notin I}}\left(  -1\right)  ^{\sum I-\left\vert I\right\vert }
\det\left(  A_{\left[  n\right]  \setminus I,\left[  d\right]  }\right)
e_{I}\wedge e_{k}$
(here, we have removed all addends which satisfy $k\in I$, because all such
addends contain the vanishing factor $e_{I}\wedge e_{k}=0$), we should obtain
$\left(  \left(  \operatorname*{Cof}A\right)  \circ A^{T}\right)  \left(
w\right)  =\left(  \operatorname*{Det}A\right)  \wedge w$. This proves Theorem 1.
