Let $\times$ be this 7-dimensional cross product and let $\hspace{.04 in}f$ be a bilinear map
on $\mathbb{R}^7$ which satisfies the orthogonality and magnitude conditions.
Does there necessarily exist a linear map $\: \phi : \mathbb{R}^7 \to \mathbb{R}^7 \:$ such that
for all vectors $x$ and $y$ in $\mathbb{R}^7\hspace{-0.03 in}$, $\;\;\; \phi \hspace{.02 in}(\hspace{.05 in}f(x,\hspace{-0.04 in}y\hspace{-0.02 in})) \: = \: \phi(x) \times \hspace{.02 in}\phi(\hspace{.03 in}y\hspace{-0.02 in}) \:\:\:\:$?
Here I show that such a $\phi$ is necessarily an isometry:
By bilinearity, $\mathbf{0} \times$ anything gives $\mathbf{0}$. $\:$ By taking a non-zero orthogonal vector and applying
the magnitude condition, one gets that for every non-zero vector there is another vector
such that their cross product is non-zero. $\:$ Thus $\operatorname{Kernel}(\phi)$ is trivial. $\:$ Since $\phi$ is linear
and $\mathbb{R}^7$ is a finite-dimensional real vector space, that means $\phi$ is a homeomorphism.
Fix a non-zero vector $x$ in $\mathbb{R}^7\hspace{-0.04 in}$, let $\: \hat{\psi} : \mathbb{R}^7 \to \mathbb{R}^7 \:$ be given by $\;\;\; \hat{\psi}\hspace{.02 in}(\hspace{.03 in}y\hspace{-0.02 in}) \: = \: x\times y \;\;\;$, $\;\;\;$ and let
$\psi : \operatorname{Range}\hspace{-0.05 in}\left(\hspace{-0.03 in}\hat{\psi}\hspace{-0.02 in}\right) \to \operatorname{Range}\hspace{-0.05 in}\left(\hspace{-0.03 in}\hat{\psi}\hspace{-0.02 in}\right) \:$ be given by $\: \psi\hspace{.03 in}(\hspace{.03 in}y\hspace{-0.02 in}) = \hspace{.02 in}\hat{\psi}\hspace{.02 in}(\hspace{.03 in}y\hspace{-0.02 in}) \;$. $\;\;\;$ By the orthogonality condition, all elements of $\operatorname{Range}\hspace{-0.05 in}\left(\hspace{-0.03 in}\hat{\psi}\hspace{-0.02 in}\right)$ are orthogonal to $x$. $\:$ Hence, by the magnitude condition, $\psi$ multiplies norms by $||\hspace{.02 in}x\hspace{.02 in}||$. $\:$ Since $x$ is non-zero, there are non-zero vectors in $\operatorname{Range}\hspace{-0.05 in}\left(\hspace{-0.03 in}\hat{\psi}\hspace{-0.02 in}\right)\hspace{-0.03 in}$. $\;\;\;$ Iterating $\hspace{.02 in}\psi \hspace{.02 in}$ will make
those vectors converge to $\mathbf{0}$ if and only if $\: ||\hspace{.02 in}x\hspace{.02 in}|| < 1 \;$. $\;\;\;$ For all non-zero vectors $x$, $||\hspace{.02 in}x\hspace{.02 in}||$ is the largest
scalar $c$ such that iterating the $\psi$ constructed from $x\hspace{.03 in}/c$ does not make the non-zero elements of
its range converge to $\mathbf{0}$. $\;\;\;$ Since $\phi$ is a linear homeomorphism, $\: \phi\hspace{.02 in}(\mathbf{0}) = \mathbf{0} \:$ and $\phi$ does not affect convergence to $\mathbf{0}$ or the lack thereof, so $\phi$ does not affect those $c$s. $\;\;\;$ Therefore $\phi$ is an isometry.
By linearity and bilinearity, $\hspace{.04 in}f$ and $\phi$ can easily be represented with finite lists of real numbers,
and for the natural such representation, the conditions on $\hspace{.04 in}f$ are equivalent to a fairly simple
conjunction of unit sphere equations and the possible-identity is equivalent to a fairly simple
conjunction of bilinear equalities in which the coefficients are all $1\hspace{-0.02 in}$, each product has one number
from $\hspace{.04 in}f\hspace{.025 in}$'s list and one number from $\phi$'s list, and ignoring the possibility of some of the variables
being outside the interval $[-1,\hspace{-0.04 in}+\hspace{-0.02 in}1]$ would not affect the truth value of the resulting sentence.
Thus, by a sort of thing I've been referring to fairly often recently,
there constructively is an algorithm that will answer my question.
However, I'm fairly confident that running an algorithm constructed like that would be wildly impractical, and am hoping that the answer is already known or can be found in some easier way.