If $T$ is a unitary linear operator in the complex space $(V, ⟨\cdot|\cdot⟩)$ is it also unitary in any other inner product space $(V, \cdot)$?

Question

I am currently reading Steinberg's "Representation Theory of Finite Groups", in which the result: Let $G$ be a finite group and $V$ a complex inner product space. Then, any representation $\varphi$ of $G$ on $V$ is equivalent to a unitary representation (i.e. a representation in which every linear operator associated with an element of $G$ is unitary) is proven. The proof consists of defining a new inner product on $V$, in which every linear operator associated with an element of $G$ is unitary, but, does this imply that the operators are unitary on V under the "original" inner product?

Answer 1

Let $(v_1, v_2)$ be any basis of $\mathbb{R}^2$, and let $J \colon \mathbb{R}^2 \to \mathbb{R}^2$ be the linear map $Jv_1 = v_2$ and $Jv_2 = -v_1$. This operator is almost certainly not orthogonal (i.e. unitary) with respect to the dot product, in fact it is orthogonal if and only if $v_1$ and $v_2$ are orthogonal vectors of the same length. However, if we define a new inner product $\langle -, - \rangle$ on $\mathbb{R}^2$ by declaring $(v_1, v_2)$ to be an orthonormal basis, then $J$ is an orthogonal transformation: in fact it looks like a 90 degree rotation.