Let $J$ be an antilinear operator on a given Hilbert space $\mathcal{H}$. The adjoint of $J$ is defined by the following formula $$\langle J\xi,\eta\rangle=\overline{\langle\xi,J^\ast\eta\rangle},\ \ \ \forall\xi,\eta\in \mathcal{H}.$$ Is it true that if $J^2=\epsilon \mathbf{1}$ with $\epsilon^2=1$, then $J^\ast=J^{-1}$ ? I know that it is true if $J$ is in addition antiunitary (which in that case means $\langle J\xi,J\eta\rangle=\overline{\langle \xi, \eta\rangle}$), but I don't see why it should be true in general (if it is false, what is the simplest counterexample?).
On $\mathbb C^2$, define $J(z_1,z_2) = (2\bar z_2, \bar z_1/2)$. Then $J^2=\mathbf 1$. Also, $J^*(z_1,z_2) = (\bar z_2/2, 2\bar z_1)$ because $$\langle Jz, w\rangle = 2\bar z_2 \bar w_1 + \frac12 \bar z_1\bar w_2$$ is consistent with $$\langle z , J^*w\rangle = \frac12 z_1 w_2 + 2z_2 w_1$$ Clearly $J^*\ne J^{-1}$ because $J^{-1}=J$.