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Suppose that $G$ is a finite group and let $U$, $V$, and $W$ be representations of $G$. Let $\chi_V$ be the character of $V$, etc. Using the natural inner product, we calculate that \begin{gather*} \langle \chi_U , \chi_V \chi_W \rangle = \langle \chi_{V^*} \chi_{U}, \chi_W \rangle = \langle \chi_W, \chi_{V^*} \chi_{U} \rangle \end{gather*} and hence \begin{align*} \text{Hom}_G(U, V \otimes W) \cong \text{Hom}_G(W, V^* \otimes U). \end{align*} Is this isomorphism natural? I haven't been able to show this using tensor-hom adjunction or similar standard isomorphisms.

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No, this is not a natural isomorphism. In particular, in the case that $G$ is trivial and $U$ and $W$ are $\mathbb{C}$, you'd be asking for a natural isomorphism between $V$ and $V^*$ for any vector space $V$.

More generally, these two vector spaces are naturally dual (assuming everything is finite-dimensional). There are natural isomorphisms of $G$-representations $$\operatorname{Hom}(U,V\otimes W)\cong U^*\otimes V \otimes W$$ and $$\operatorname{Hom}(W,V^*\otimes U)\cong W^*\otimes V^*\otimes U\cong (U^*\otimes V\otimes W)^*.$$

Writing $X=U^*\otimes V\otimes W$, we see that $\operatorname{Hom}_G(U,V\otimes W)$ and $\operatorname{Hom}_G(W,V^*\otimes U)$ are naturally identified with the invariants $X^G$ and $(X^*)^G$, respectively. But in fact, $X^G$ and $(X^*)^G$ are naturally dual, since $X$ naturally splits as a direct sum $X^G\oplus Y$ where $Y$ is the sum of all nontrivial irreducible subrepresentations of $X$ and then $(X^*)^G$ is just the direct summand $(X^G)^*$ in the dual direct sum decomposition of $X^*$.

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