$\newcommand\norm[1]{\lVert#1\rVert}$Here is a simple argument showing that, with the above notations, if $U$ is a unitary operator, and $\tilde{U}$ an antiunitary operator in $\mathbf{C}^n$, one always has $\norm{U-\tilde{U}}\ge \sqrt{2}$.
We note that, by definition of the operator norm we have $\norm{U-\tilde{U}}=\sup_{\norm z=1} \norm{Uz-\tilde{U}z}=\sup_{\norm z=1}\norm{z-U^{-1}\tilde{U}z}$. The operator $U^{-1}\tilde{U}$ is antiunitary, hence of the form $z\mapsto U_0\bar{z}$, where $U_0$ is some unitary operator in $\mathbf{C}^n$. Hence we must show that for any such $U_0$ we have $\sup_{\norm z=1}\norm{z-U_0\bar{z}}^2\ge 2$. Now $U_0$ is diagonalizable, and its eigenvalues have unit modulus. Let $u_0$ be any normalized eigenvector, and let $e^{i\phi_0}$ be its eigenvalue. Setting $z=e^{i\alpha}\bar{u}_0$, we have $\norm{z-U_0\bar{z}}^2=\norm{e^{i\alpha}\bar{u}_0-e^{i(\phi_0-\alpha)}u_0}^2=\norm{\bar{u}_0-e^{i(\phi_0-2\alpha)}u_0}^2$ $=2-2\Re\{e^{i(\phi_0-2\alpha)}(\bar{u}_0|u_0)\}$.
It follows that $\norm{U-\tilde{U}}^2\ge \sup_{\alpha\in [0,2\pi]}(2-\Re\{e^{i\alpha}(\bar{u_0}|u_0)\})\ge 2$, Q.E.D.
I suspect one could do better (for example when $n=1$ one finds easily that always $\norm{U-\tilde{U}}=2$), but this is not the point here.
