I want to make sure I completely understand the isomorphism classes of smooth unitary irreducible finite-dimensional representations of $U(n)$. We have the irreducible defining representation $R$, and we can apply any Young diagram to this, hitting it with a Schur functor, to get a load more irreps. We can also take the complex conjugate $R^*$ of the defining representation, and hit this with all the Young diagrams, to get another series of irreps. Some of these irreps will actually be the same, if the number of rows in our Young diagrams gets as large as $n$, but I'm not worried about that.
I understand how to tensor together representations that arise from applying Young diagrams to $R$, and decompose this into a direct sum of irreducible representations. But what if I tensor together an irrep coming from a Young diagram applied to $R$, and an irrep coming from a Young diagram applied to $R^\star$? How does this product decompose into a sum of irreps? I've convinced myself it just isn't possible when you're restricted to using the irreps that arise from applying Young diagrams to $R$ and $R^\star$ --- there must be some more of them. For example, consider $U(2)$ and the representation $R \otimes R^\star$, where $R$ is the defining representation. Since $R$ and $R^\star$ are dual, this must decompose as $R \otimes R^* \simeq 1 \oplus X$, where $X$ is a 3-dimensional self-dual possibly-reducible representation. But I don't think it's possible to build such an $X$ by taking direct sums of the irreps described in the first paragraph.
So, apart from applying Young diagrams and conjugate Young diagrams to the defining representation, how to you get the rest of the of the irreps? In particular, I don't think I know how to build any self-dual irreps, apart from the trivial irrep, but some have to exist by the argument in the previous paragraph.