For a $d$-box Young diagram $\lambda$, the Schur functor is a functor $S_\lambda: \text{Vect}\rightarrow \text{Vect}$. If $\lambda = d$ then $S_\lambda V=S^d V$ the $d$-th symmetric power of $V$, whilst if $\lambda=(1,\dots,1) = (1^d)$ we have $S_\lambda V=\bigwedge^d V$, the $d$-th exterior power.
Suppose $V$ is additionally an irreducible representation of the symmetric group $S_n$. Such representations can be indexed by an $n$-box Young diagram, so $V=V_\mu$ for some such diagram $\mu$. For instance the standard representation is labelled by $\mu=(n-1,1)$, and a well-known result says that $\bigwedge^d V_{(n-1,1)}=V_{(n-d, d)}$. In terms of Schur functors, $$S_{(1^d)}(V_{(n-1,1)})=V_{(n-d,d)}$$ Are there any known generalisations of this equation? I.e. any rules for decomposing $S_\lambda(V_\mu)$ as a direct sum of irreducibles of $S_n$ for general $\mu,\lambda$? I'm hoping for a rule in terms of Young diagrams, that takes a $d$-box $\lambda$ and $n$-box $\mu$, and spits out a set of $n$-box diagrams $\{\mu_1,\dots,\mu_k\}$ such that $S_\lambda(V_\mu)=\bigoplus_i V_{\mu_i}$, but I somehow doubt this exists.