As in RW's answer, I assume $X$ is compact, and $C(X)$ means continuous real-valued functions (the extension to complex continuous functions is routine). But now assume something much stronger: $X$ is zero-dimensional.

Then, modulo coboundaries (functions of the form $f - T\circ f$), and assuming something like recurrence (or some condition known to guarantee an obvious pre-ordering is a valid partial ordering, see below), these are just the functions that are infinitesimals plus a constant function; simply note that $f- \int f\,d\mu$ (for these special $f$s) vanishes at all the measures (or more accurately, vanishes at all the positive linear functionals on $C(X)$ that are induced by the invariant measures). This describes precisely what are known as the infinitesimal subgroup of the (pre-) ordered group $C(X)/(\text{image}(I-T))$. [See towards the end of

M Michael Boyle and David E Handelman
*Orbit equivalence, flow equivalence and ordered cohomology*
Israel Journal of Mathematics
December 1996, Volume 95, Issue 1, pp 169-210.]

Most of this deals with integer-valued continuous functions---which of course are interesting only for highly disconnected spaces; but at one point we consider real-valued functions. Somewhat later, it was shown by Nick Ormes that the ordered group (here a vector space) is generically laced with infinitesimals.]

Without something like recurrence, the pre-ordering is not likely to be a proper partial ordering, and in that case the answer still consists of the infinitesimals, but with a slightly different interpretation.