BRST transformation of adjoint spinor in Yang-Mills-Theory with matter fields a dirac spinor $\psi$ transforms under BRST as
$$\psi \to \delta_\Omega\psi=i\eta\psi $$
with $\eta$ being a ghost field. If I want to get the transformation of the adjoint spinor $\bar \psi$ I get by using the invariance of $\bar \psi \psi$ 
$$ 0=\delta_\Omega (\bar \psi\psi)=(\delta_\Omega\bar\psi)\psi - \bar\psi (\delta_\Omega\psi) \quad \Rightarrow \quad \delta_\Omega \bar\psi=i\bar\psi \eta$$
If I now want to get the transform directly, I get
$$ \delta_\Omega \psi^\dagger \gamma_0=[\psi^\dagger,\Omega]_+\gamma_0=([\psi,\Omega]_+)^\dagger \gamma_0=(i\eta\psi)^\dagger\gamma_0=-i\bar\psi\eta $$
So I get different results. Where is my error? What I am not sure about, is whether if I have $(\eta\psi)^\dagger$ if this is equal to $\psi^\dagger\eta$ or $-\psi^\dagger \eta$ as the transpose part should be purely in the dirac space.
Thanks in advance.
 A: The reason for the discrepancy is that the BRST operator is not self adjoint $\Omega^{\dagger} \ne \Omega$, This can easily be seen from its action on the ghosts:
$$\delta c^{a} = \epsilon \{\Omega, c^a\} = i \epsilon f^{a}_{bc} c^b c^c$$
$$\delta \bar{c}^{a} = \epsilon \{\Omega, \bar{c}^a\} = - i \epsilon b^a$$
$$\delta b^{a} = \epsilon \{\Omega, b^a\} = 0$$
(It is clear that these equations cannot be obtained by simple Hermitian conjugation). 
Furthermore the operator $\Delta = \{ \Omega, \Omega^{\dagger}\}$ plays an important role in the BRST theory, please see for example Aspects of BRST quantization   by J.W. van Holten. The zero modes of the BRST Laplacian can be chosen as a basis for the physical states.
Please observe that in the (physical) fermion sector, the BRST operator is also not self adjoint, because both its actions on the spinor and its Dirac adjoint are proportional to the ghost and not to its conjugate (which appears only in the ghost kinetic term. 
Also, please notice that both actions do not contain the imaginary unit
($ i =  \sqrt{-1}$), since the transformation must preserve the reality of the Lagrangian ($\epsilon$ is a real Grassmann variable).
$$\delta \psi^i = \epsilon \{\Omega,  \psi_i\} = \epsilon  c^a (T_a)_j^i \psi^j$$
$$\delta \bar{\psi}_i = \epsilon \{\Omega, \bar{\psi}_i\} = \epsilon \bar{\psi}_j (T_a)^j_i  c^a$$
A: In your case, I think that $\eta$ is playing the role of $\epsilon$ in David Bar Moshe 's answer, that is : $\eta$ is a real Lorentz-scalar Grassmann variable. So, you will have :
$$\delta_\Omega \bar\psi= \overline{\delta_\Omega\psi} = \overline {i\eta\psi} = -i \eta  \bar\psi = i  \bar\psi \eta$$
