In this popular 2019 MO question, user მამუკა ჯიბლაძე asked:
The spaces $\operatorname{S}^2(k^n)$ and $\Lambda^2(k^{n+1})$ from the title have equal dimensions. Is there a natural isomorphism between them?
The answer was affirmative: let $E=k^2$, then the Wronskian isomorphism gives an identification $\bigwedge^m(S^{m+r-1}(E)) \rightarrow S_m(S^r(E))$ as $\operatorname{SL}(E)$-modules.
(Here, $S_m(V)=(V^{\otimes m})^{S_m}$ and $S^m(V)$ is the symmetric power, as a quotient of $V^{\otimes m}$.)
The recent paper Modular plethystic isomorphisms for two-dimensional linear groups of McDowell–Wildon contains the proof of this over any field.
My question is in a similar vein. Let $\nabla^\lambda$ be the Schur functor of the partition $\lambda$, and consider $\lambda_0=(2,1)$. Then, the $\operatorname{SL}(E)$-character of $\nabla^{\lambda_0}(S^m(E))$ is simply $$ \newcommand\qbinom{\genfrac[]0{}}[2]\qbinom{m+2}3, $$where the brackets indicate that these are quantum numbers and binomials, so for instance $[2]=q+q^{-1}$.
Notice that this holds for any $m$. This suggests the possibility that there exists a natural isomorphism $\nabla^{\lambda_0}(S^m(E))\cong E\otimes V_m$, for some "uniformly defined" $\mathrm{SL}(E)$-module $V_m$ of dimension $\qbinom{m+2}3$.
Is there such a natural isomorphism?