# Pre-metric electrodynamics: additional symmetry of the tensor density

I am currently reading this paper. How can I prove (eq. 7 of the paper) that if we assume

$$\chi ^{abcd} = \chi ^{cdab}$$

$$F \wedge \# F' = F' \wedge \# F$$

and vice versa?

$$F$$ is the electromagnetic 2-form and the Maxwell equations are given by

$$d F = 0 \quad , \quad d \# F =J$$

where

$$(\# F)_{ab} = \frac{1}{2} {\kappa _{ab}}^{cd}F_{cd} = \frac{1}{4}\epsilon_{abcd}\chi^{cdef}F_{ef}$$

where $$\epsilon_{abcd}$$ is the Levi-Civita symbol and

$$\chi^{cdef} = \frac{1}{2}\epsilon ^{abcd} {\kappa_{cd}}^{ef}$$

with

$${\kappa_{ab}}^{cd} = - {\kappa_{ba}}^{cd} = - {\kappa_{ab}}^{dc} \quad , \quad \chi^{abcd} = -\chi^{bacd} = -\chi^{abdc}.$$