Is it true that every involution $\sigma$ (i.e., $\sigma^2=identity$) of an Enriques surface $X$ acts trivially on $K_X^{\otimes 2}$ i.e., for any $\omega\in K_X^{\otimes 2}$ we have $\sigma^* \omega=\omega$, where by $K_X^{\otimes 2}$ we mean the tensor 2 of the conanical bundle of $X$.
Let me try an argument different from Christian's: $\sigma$ does not act freely as $\chi(\mathcal O_X)=1$ and hence not divisible by $2$. At a fixed point $x$, $\sigma$ acts by $\pm1$ on the fibre of $\omega_X$ and hence acts by $1$ on the fibre of $\omega_X^{\otimes2}$. It also acts by a scalar on a global nonzero section of $\omega_X^{\otimes2}$ but as that section is nonzero at $x$ this scalar must be $1$. Addendum: It seems that it is essential that there are fixed points. If we look at a bielliptic example; $E\times F$ the product of two elliptic curves with $\tau$ acting by an automorphism of order $4$ on $E$ (assumed to have one) and translation by an element of order $4$ on $F$ then if we divide by $\tau^2$ we have that $\tau$ induces an involution which acts by multiplication by $1$ on global sections of $\omega^{\otimes 2}$. 


Yes! However, as Rita, Torsten and Ru pointed out, my first ideas were too simpleminded. Although this makes the remarks by them somewhat unreadable, let me give the corrected answer: So, let $X$ be a complex Enriques surface and $\sigma$ an involution. Let us denote by $\tilde{X}\to X$ the associated K3cover, and let $\tau$ be the associated involution, i.e., $X=\tilde{X}/\langle \tau\rangle$. Now, the automorphism group of $X$ in terms of $\tilde{X}$ is $$ {\rm Aut}(X)={\rm Aut}(\tilde{X},\tau) := {} ( \psi\in{\rm Aut}(\tilde{X}) \psi\tau=\tau\psi ) / \langle\tau\rangle $$ (typesetting the usual brackets does not seem to work?!). In particular, $\sigma$ lifts to an automorphism $\tilde{\sigma}$ of $\tilde{X}$. Clearly, we have $\tilde{\sigma}^2\in\langle\tau\rangle$. I claim that $\tilde{\sigma}^2=\{\rm id}$. For otherwise, we would have $\tilde{\sigma}^2=\tau$, and $\tilde{\sigma}$ would be an automorphism of order $4$. In this case, since $\tau$ acts freely on $\tilde{X}$, the same would be true for $\tilde{\sigma}$. However, a K3 surface cannot possess a fixedpoint free automorphism of order $4$: the quotient $S$ of $X$ by this automorphism would satisfy $\chi({{\mathcal O}_S})=1/2$, which is absurd. Thus, $\tilde{\sigma}$ is an involution on $\tilde{X}$. Being an involution, $\tilde{\sigma}$ acts as $\pm{\rm id}$ on the $1$dimensional vectorspace $H^0(\omega_{\tilde{X}})$. Since $H^0(\omega_{\tilde{X}})^{\otimes2}\to H^0(\omega_{\tilde{X}}^{\otimes 2})$ is onto, we conclude that $\tilde{\sigma}$ acts trivially on global sections of $\omega_{\tilde{X}}^{\otimes2}$. Now, $\tilde{\sigma}$ induces $\sigma$ on $X$, and global sections of $\omega_X^{\otimes2}$ pull back to global sections of $\omega_{\tilde{X}}^{\otimes2}$. Since $\tilde{\sigma}$ acts trivially on these, we conclude that $\sigma$ acts trivially on global sections of $\omega_X^{\otimes2}$. It is less obvious, but still true, that automorphisms of order $3$ and $5$ also act trivially on global sections of $\omega_X^{\otimes2}$. Mukai and Ohashi exploit this in their recent analysis of automorphisms of Enriques surfaces. 

