Let $X$ be a scheme, and let $\sigma : X \rightarrow X$ be an involution of $X$ (so $\sigma^2 = id$). I can always get an induced involution on the structure sheaf of the affinization of $X$ by noting that $LR(\mathcal O_X) = LR(\sigma_* \mathcal O_X)$, where $R$ is global sections and $L$ is the associated sheaf functor. However, if I want to turn this into an involution of $\mathcal O_X$, I need to conjugate by the counit $\epsilon : LR \rightarrow id$ of the adjunction, which is not an natural isomorphism unless $X$ is affine.
I think a potential counterexample is $\mathbb P^1$ with involution defined by $[x:y] \mapsto [y:x]$, in which case the problem is that there's not an open cover of affines which are invariant under the involution. It seems like one would want to send $x \mapsto \frac{1}{x}$ on each of the affine covers, which is impossible.
However, even if there is such an invariant cover, I was struggling to glue some locally defined maps -- how would one show that $\epsilon_{\mathcal O_{U_i}} LR(\sigma^\# |_{U_i})\epsilon_{\mathcal O_{U_i}}^{-1} $ and $\epsilon_{\mathcal O_{U_j}} LR(\sigma^\# |_{U_j})\epsilon_{\mathcal O_{U_j}}^{-1} $ restrict to the same map on $U_{ij}$?
Edit: As pointed out in the comments, let me only ask for an additive involution on the sheaf.
An involution of a scheme does not naturally induce any sort of involution of its structure sheaf, even in the affine case. Your idea is essentially that when $X=\operatorname{Spec} A$, $\sigma$ induces an involution of $A$ which then corresponds to an involution of $\mathcal{O}_X$. But this is incorrect, because the involution of $A$ is an involution of rings, not an involution of $A$-modules. To get a corresponding homomorphism of sheaves, you would need a homomorphism of $A$-modules.
Geometrically, this should not be surprising. An involution of $X$ can move points of $X$, whereas an involution of $\mathcal{O}_X$ has to be a sheaf morphism and so must keep every point of $X$ fixed in some sense. Concretely, an involution of $\mathcal{O}_X$ as an $\mathcal{O}_X$-module is given by multiplication by some $f\in A$ such that $f^2=1$. It would be very weird to naturally be able to obtain such an $f$ from an involution of $A$ as a ring.