A conjugate LT in an inner product space

Question

Let $V$ be a finite dimensional complex inner product space. Let $J$ be a conjugate linear map from $V$ to $V$ such that $J^2 =1$.

Can we say $\langle Ju, Jv \rangle = \langle v, u \rangle$ for all $u, v$ in $V$?

Conjugate linear means $J(u+v) = J(u) + J(v)$ and $J(a.v) = \overline{a}.J(v)$

Answer 1

It is in general wrong. Consider on $V=\mathbb C^2$ the antilinear mapping $$ J(z) = \underbrace{\begin{bmatrix} 1 & i \\ 0 & 1 \end{bmatrix}}_{=:A} \bar z. $$ Then, we have $$ J^2(z) = A\overline{(A \bar z)} = A\bar A z = \begin{bmatrix} 1 & -i + i \\ 0 & 1\end{bmatrix} z = z$$ and $$ \langle J e_1, Je_2 \rangle = (A \bar e_1)^H (A \bar e_2) = (A e_1)^H (A e_2) = 1\cdot i + 0\cdot 1 = i \ne 0 = \langle e_2, e_1 \rangle. $$

As asides:

1. Every antilinear mapping on $\mathbb C^n$ looks like $z\mapsto A\bar z$.
2. $A$ is symmetric(!) if and only if $J$ has the mentioned property: $$ \langle J(u), J(v) \rangle = \langle A\bar u, A \bar v \rangle = \langle \bar u, A^H A \bar v \rangle = \langle \bar u, \bar v \rangle = \langle v, u \rangle $$ holds for every $u,v$ if and only if $A^H A = I$. That is $\bar A = A^{-1} = A^H = \bar A^T$. That is, $A$ is symmetric.