Is it true that $$\langle Au,Av\rangle = \langle u,A^{*}Av \rangle$$
for some inner product space $V,$ vectors $u,v$ and a square matrix $A$?
For the standard inner product, this is trivially true if $A$ is Hermitian, but what about other cases?
What if the inner product is not standard?
What if $A$ is not Hermitian?
When does the above hold true?
$A*$ refers to the conjugate transpose of $A.$
As mentioned in the comments, on a finite dimensional space $V$ you have essentially the euclidean inner product. This means that any inner product on $V$ can be written as
$$ \langle u, v \rangle_M = u^\ast M v $$
for a hermitian positive definite matrix $M$ (see e.g. here). But $(V, \langle \cdot, \cdot \rangle_M)$ is isometrically isomorphic to $(V, \langle \cdot, \cdot \rangle_E)$, where $\langle \cdot, \cdot \rangle_E$ denotes the euclidean inner product. To see this note that $M$ can be written as a square of a hermitian matrix $M = N \cdot N.$ Then $$ (V, \langle \cdot, \cdot \rangle_M) \to (V, \langle \cdot, \cdot \rangle_E); \quad x \mapsto Nx $$ gives an isometric isomporphism, as $$ \langle u, v\rangle_M = u^\ast M v = u^\ast NN v = \langle Nu, Nv \rangle_E. $$
Now we show the desired property for the euclidean inner product, which is
$$ \langle u, v \rangle = u^\ast v $$ for arbitrary $u, v \in V.$
Thus \begin{align*} \langle Au, Av \rangle = (Au)^\ast Av = u^\ast A^\ast Av = \langle u, A^\ast Av \rangle. \end{align*}