TITLE: Is there a Kähler manifold with no anti-holomophic involution?
QUESTION [10 upvotes]: That is, is there a Kähler manifold $X$ on which there is no map
$$
\tau:X\to X
$$
such that
$$
d\tau\circ I=-I\circ d\tau
$$
and
$$
\tau\circ \tau=\mathrm{Id}_X?
$$

REPLY [8 votes]: There even exists a (compact) Kähler manifold with no anti-holomorphic self-diffeomorphism at all. Namely a suitable elliptic curve.
Indeed, consider $\mathbf{C}/\Lambda$, for some lattice $\Lambda$. Let $f$ be an anti-holomorphic self-diffeomorphism. Up to compose with a translation, we can suppose that $f(0)=0$. Let $F$ be the unique lift $\mathbf{C}\to\mathbf{C}$ mapping $0$ to $0$. Then $F$ is a self-covering, and hence is a diffeomorphism, and commutes with $\Lambda$-translations. Write $G(z)=\overline{F(z)}$. Then $G$ is a holomorphic self-diffeomorphism of $\mathbf{C}$ and fixes $0$, hence $G(z)=\bar{q}z$ for some $q\in\mathbf{C}^*$. Since $G$ maps $\Lambda$ to $\overline{\Lambda}$ which has the same covolume, we have $|q|=1$. We have $F(z)=q\bar{z}$ for all $z$, and $F(\Lambda)=\Lambda$. So $\Lambda=q\overline{\Lambda}$. That is, $\Lambda$ is preserved by some reflection $z\mapsto q\bar{z}$.
Next is it enough to see that "most" lattices are not preserved by any reflection. Indeed, we can choose coordinates so that $\Lambda$ has the basis $(1,z)$ with $z$ belonging to the usual "strip" ($|z|\ge 1$, $\frac12<\mathrm{Re}(z)\le \frac12$, $\mathrm{Im}(z)>0$). Then such a lattice is preserved by a reflection if and only if $\mathrm{Re}(z)\in\{0,1/2\}$. (The latter point was already observed in the previous answer using another language, but the easy argument could easily be explained.)
(Indeed we see in these cases that there exists a anti-holomorphic self-diffeomorphism if and only if there is one that is in addition involutive, and furthermore preserves the standard flat Kähler structure.)