Is the highest weight space of $V^{\otimes k}$ an irreducible representation of $S_k$? Let $V$ be $n$-dimensional vector space over $\mathbb{C}$ with given action of $GL_n$. Let $W_\lambda$ be the highest weight space of weight $\lambda$ in $V^{\otimes k}$. Is $W_\lambda$ an irreducible $S_k$ representation?
This should follow from Schur-Weyl duality, but I don't see why.
 A: Let $\mu$ be the highest weight of $V$. It should be clear that the highest weight space of $V^{\otimes k}$ is $V^{\otimes k}_\mu$ (so $\lambda = k\mu$). This is 1-dimensional and so irreducible. No need for Schur-Weyl duality except to clarify that this is preserved under the action of $S_k$.
Edit:
To make it even clearer why the highest weight space is $V^{\otimes k}_\mu$ consider the following: $V = \bigoplus_{\omega} V_\omega$ so $V^{\otimes k} = \bigoplus_{\omega_1,\dots,\omega_k} V_{\omega_1}\otimes \cdots \otimes V_{\omega_k}$ where $\omega, \omega_1,\dots,\omega_k $ range over all the weights of $V$. Note that $V_{\omega_1}\otimes \cdots \otimes V_{\omega_k} \leq W_{\omega_1+\cdots+\omega_k}$. Then since all of our summands live in specific weights spaces and they span the whole space we can easily conclude that:
$$W_\nu = \bigoplus_{\omega_1+\cdots+\omega_k = \nu}V_{\omega_1}\otimes \cdots \otimes V_{\omega_k} $$
The highest possible weight that we see this way is $ \lambda = k\mu $ and the only one of the summands which has that weight is $V^{\otimes k}_\mu$.
While there are are multiple irreducible subrepresentations each with its own highest weight these weights are different in general and thus $S_k$ does not move between these since its action preserves weight spaces. Thus the span of all of these weight spaces is certainly not an irreducible representation.
Edit2:
Note in general that a non-irreducible representation doesn't have to have a highest weight but this one does. It is not too hard to see that $(\omega_1+\cdots+\omega_k) \leq k\mu $ since $ \omega\leq \mu$ for all weights of $V$.
Edit3:
From some of your comments I think I'm starting to see where your confusion arises. $\lambda$ is commonly used as notation for weights but also for partitions. These are not the same things however. Schur-Weyl is usually written in terms of partitions:
$$ V^{\otimes k} = \bigoplus_{|\lambda|=k} V_\lambda \otimes W_\lambda$$
Here $W_\lambda$ is an irreducible $S_k$-module corresponding to the partition $\lambda$. Meanwhile $ V_\lambda $ is an irreducible $GL(n)$-module although we may need to think a little harder to find out what $V_\lambda$ is since we haven't been given its highest weight (at least not explicitly). Together $V_\lambda \otimes W_\lambda$ is called an isotypic component. The highest weight space of $V_\lambda$ is 1 dimensional so the highest weight space of $V_\lambda \otimes W_\lambda$ is isomorphic to $ W_\lambda$ and is an irreducible $S_k$-module (always - no condition on the partition needed). Of course the highest weight in one component may appear as a non-highest weight in another component but your stated condition ($\lambda$ having at most $n$ parts) certainly doesn't pick out the cases where this doesn't happen. For example, the partition $(1,1,\dots,1)$ corresponds to the symmetric power $S^k V$ which is where the highest weight I mention in my first answer lives. In that case the whole weight space is irreducible since it is 1-dimensional as mentioned. Meanwhile $(\mathbb{C}^2)^{\otimes 4}$ has a submodule corresponding to the partition $\{2,2\}$ which is trivial as a $GL(n)$-module so the highest weight is $0$ and this weight is found in every other component and so the $0$ weight space is certainly not irreducible as an $S_k$-module.
