$T: \mathbb R^n \to \mathbb R^n$, $\langle Tu,v\rangle=\langle u,T^*v\rangle$, is $T^*=T^t$ regardless of inner product? Basic question in linear algebra here. $T$ is a linear transform from $\mathbb R^n$ to $\mathbb R^n$ defined by $T(v)=Av$, $A\in \mathrm{Mat}_n(\mathbb R)$. We are given some inner product $\langle ,\rangle$ of $\mathbb R^n$. Does not have to be the standard one, just some random inner product.
let $T^*$ be a linear transform from $\mathbb R^n$ to $\mathbb R^n$ such that for all $u,v \in \mathbb R^n$: $\langle T(u),v\rangle=\langle u,T^*(v)\rangle$
I know that if $\langle ,\rangle$ is the standard inner product of $\mathbb R^n$, then $T^*(v)=A^tv$. My question is, does this hold for all inner products?
If it isn't, given some inner product and the transform $T$, how can I find $T^*$?
And if it is true, then why?

No it does not hold for any inner product.
It is hard hard to show that any other inner product $\langle\cdot,\cdot\rangle_*$ can be represented as
$$
\langle x,y\rangle_\star=\langle x,Sy\rangle,
$$
where $S$ is a positive definite matrix. So
$$
\langle Tx,y\rangle_\star=\langle x,T^*Sy\rangle \ne \langle x,ST^*y\rangle=\langle x,T^*y\rangle_\star,
$$
unless $ST^*=T^*S$.
