Let $\omega$ be a differential form on $S^1.$ There is an action of a circle on itself by rotations ($A: S^1 \to S^1$ as $A(x)=A+x$ if we think of a circle as an additive group). Let's define the differential form $A(\omega)$ which is the "averaging" (namely, $A(\omega)$ is invariant with respect to the action $S^1$) of the form $\omega$ as $A(\omega_p)(x)= \int_{S^1} A^{*} \omega_p (x) dA.$
My question is: how to realize that $A(\omega)$ and $\omega$ are cohomologous. I think that's because of connectedness of a circle but I can't find such $f$ that $df=\omega-A(\omega).$