Answer: There is none.

The issue at hand is that the Kaehler invariance is just that - an invariance, not a continuous symmetry of the fields. Most prominently the superpotential must transform as
$$ \mathcal W \to \mathcal W e^{-h} $$
A general superpotential that leads to consistent theories is
$$ \mathcal W =\frac{1}{2} m_{ab} \phi^a \phi^b + \frac{1}{3} Y_{abc} \phi^a \phi^b \phi^c $$
with at least one of the $m_{ab}$ **and** $Y_{abc}$ non-zero. From this is is obvious, that no transformation of the *fields* $\phi^a$ exists, such that $\mathcal W \to \mathcal W e^{-h}$ without redefining the couplings.

Thus, there is a Kaehler invariance, which involves a redefinition of the couplings and has its value on its own (e.g. on non-simply connected internal spaces, the Kaehler potential might only be defined locally, with definitions on different charts being equal up to Kaehler transformations $\mathcal K' = \mathcal K + f(\phi) + \bar f(\bar\phi)$), but this is not a *symmetry* in the sense of Noether's theorem.

This post imported from StackExchange Physics at 2014-03-07 13:18 (UCT), posted by SE-user Neuneck