This question occurred to me while thinking on another one here, Name for an operation on matrices?

Can one define *in an invariant way* a binary operation on finite-dimensional vector spaces - let us denote it somehow suggestively by $(V,W)\mapsto V^{\otimes W}$ - with the property$$\dim(V^{\otimes W})=\dim(V)^{\dim(W)}?$$

Added later (and I should do it from the beginning as it is important): the construction from the linked question suggests that this operation seemingly should act on linear operators by assigning to $f:V\to X$ and $g:Y\to W$ certain operator (explicitly given there in terms of chosen bases) $$ f\dagger g:V^{\otimes W}\otimes Y\to X\otimes W, $$ with $\operatorname{rank}(f\dagger g)\geqslant\operatorname{rank}(f)\operatorname{rank}(g)$.

I'm aware that this looks even more impossible, but anyway - with bases it is done in that question.

faithfully) is in the comment above. Anyway, also note that if you want functoriality in $W$ with respect to non-invertible morphisms then there are generally no nontrivial $GL(V)$-equivariant maps $V \to V^{\otimes 2}$ or $V^{\otimes 2} \to V$, etc. In that exterior algebra observation the "base" of the exponential is not a vector space but a graded vector space, so my comment above doesn't apply. $\endgroup$differentfields?? $\endgroup$6more comments