Let $V$ be a vector space over $F$. Consider $End(V)$ as $H = GL(V) \times GL(V)$ module, via conjugation. ($(A,B)$ sends $S \in End(V)$ to $ASB^{-1}$.)
$End(V)$ is irreducible over $H = GL(V) \times GL(V)$, since it is tensor product of two irreducible $GL(V)$ modules. I want to know about the isotypic decomposition of the coordinate ring $S^*(End(V)^*)$, because of the motivation below. It also seems like a natural thing to think about. Hints would be welcome, or references.
I would also be happy to know whether or not isomorphic irreducibles can appear as submodules of $S^k(End(V))$ and $S^j(End(V))$ for different $k$ and $j$. This is an easier question and would also solve the problem in my motivation.
Thoughts: $S^k(V^* \otimes V)$ we can rewrite as $((V^*)^{\otimes k} \otimes V^{\otimes k})^{S_k}$.
Inside of this is $S^k(V^*) \otimes S^k(V)$, which is irreducible over $GL(V) \times GL(V)$. These are just the tensors which are (independently) invariant in each of the first $k$ and last $k$ factors,so there is more in $S^k(End(V))$. (The invariant tensor $f_1 \otimes f_2 \otimes v_1 \otimes v_2 + f_2 \otimes f_1 \otimes v_2 \otimes v_1$ is not in $S^2(V^*) \otimes S^2(V)$, for example.) (Or one could do a quick dimension count.)
I guess that the remaining irreducible submodules in $S^k(V^*)$ have to do with the way that $S_k$ invariance can be spread over the two factors. I don't know how to make this precise though.
Motivation: I'm trying to show that an irreducible polynomial representation of $GL(V)$ on a space $W$ has matrix coefficients which are homogeneous forms, all of the same degree. So I am looking at the map $End(W)^* \to S^*(End(V)^*)$, induced by pullback of the induced map $End(V) \to End(W)$ (my representation is polynomial). $End(W)^*$ is an irreducible $GL(V) \times GL(V)$ module, and this pullback map is $GL(V) \times GL(V)$ equivariant (well, I think so, I did the computation but maybe I'm wrong - in general, it seems that if $\psi : G \to H$ is a homomorphism of affine algebraic groups, then for the induced $G \times G$ actions on the coordinate rings, by left and right multiplication on $H$, the pullback map is $G \times G$ equivariant. The proof of equivariance just amounts to the observation that $f \psi(g_1 g g^{-1}_2) = f ( \psi(g_1) \psi(g) \psi(g_2)^{-1})$.)
I would like to show that each isotypic component of $S^*(End(V))$ lives some $S^k(End(V))$, or that if $k \not = j$, $S^k(End(V))$ and $S^j(End(V))$ have no irreducibles in common. That would force $\phi$ to land in some $S^k(End(V)^*$, which would prove that the matrix coefficients are all homogeneous of degree $k$. (I guess that there is an easier proof of this exercise, and I'd be open to a suggestion.)
If you instead consider $S^*(V \otimes W)$ as a $GL(V)\times GL(W)$ module, the thing to google is "Howe duality". Understanding the representation you want should just be a special case where $W=V^*$.
As for your question of whether or not isomorphic irreducibles can appear in different symmetric powers the answer is no. To see this it is enough to look at it as a module just over one copy of $GL(V)$, let's say the left one.
This copy of $GL(V)$ is only acting on the left factor of $V \otimes V^*$, and in particular as a representation of this copy of $GL(V)$ this is just a direct sum of $\dim(V)$ copies of $V$. But now it's clear that $S^k(V\otimes V^*)$ is a degree $k$ representation of this copy of $GL(V)$, and hence $S^k(V\otimes V^*)$ and $S^j(V\otimes V^*)$ have no common factors.