# Natural isomorphism between certain hom spaces of reprentations?

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.

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 $$X^*$$.