This answer concerns complex projective manifolds (and more generally Kahler ones).
Statement. If $X$ is Kahler then $\mathbb Z[1/2]$ action on $X$ extends to an action of $\mathbb R$ on $X$. This follows immediately from Lieberman's theorem, which I will state now (D. I. Lieberman. Compactness of the Chow scheme: applications to automorphisms and deformations of Kahler manifolds, 1978).
Denote by $Aut_0(X)$ the connected component of identity map in the group of automorphisms of $X$.
Lieberman's Theorem. Consider the action of $Aut(X)$ on $H^*(X,\mathbb Z)$,
$\phi: Aut(X)\to GL(H^*(X,\mathbb Z))$.
Then the group $Aut_0(X)$ has finite index in $ker(\phi)$.
Proof of the statement. Clearly if $\mathbb Z[1/2]$ belongs to $Aut(X)$, then it belong to $ker(\phi)$. So by Lieberman's theorem it belongs to $Aut_0(X)$. But $Aut_0(X)$ is a Lie group. This finishes the proof.
I don't know if the same reasoning can work in the real case (one might first look closer into the proof of Liebermann's result).