Hermitian conjugate of quasi-primary operator in 2D CFT

Question

In all of the sources I could find, the hermitian conjugate of a quasi-primary operator $\mathcal{O}$ with weights $(h,\bar{h})$ in (2D Euclidean) CFT is given as

$$\mathcal{O}(z,\bar{z})^\dagger = \bar{z}^{-2h} z^{-2\bar{h}} \mathcal{O}\left( \frac{1}{\bar{z}},\frac{1}{z} \right)$$

The justification is then, that we require operators to be self-adjoint in Lorentzian signature. The Euclidean time $x^0 = it$ is then inverted by complex conjugation, such that the coordinate $z = e^{x^0 + ix^1}$ is effectively inverted: $z \mapsto 1/\bar{z}$. But how does this justify the prefactors?

Since inversion is not an orientation preserving transformation (meaning $\partial_z (1/\bar{z}) = 0$), the usual formula for conformal transformation of a quasi primary field

$$\mathcal{O}(z,\bar{z}) \mapsto \left( \frac{\partial f(z)}{\partial z} \right)^h \left( \frac{\partial \bar{f}(\bar{z})}{\partial \bar{z}} \right)^{\bar{h}} \mathcal{O}(f(z),\bar{f}(\bar{z}))$$

does not apply directly. However I would guess that one could compose such a transformation with a parity transformation (which I will take as complex conjugation (?)) to get the same effect. That is, first transform $z \mapsto 1/z$ and then perform an overall complex conjugation.

The first part would look like

$$\mathcal{O}(z,\bar{z}) \mapsto \left( -\frac{1}{z^2} \right)^h \left( -\frac{1}{\bar{z}^2} \right)^{\bar{h}} \mathcal{O}\left( \frac{1}{z}, \frac{1}{\bar{z}} \right)$$

and performing an overall complex conjugation would then result in 

$$\mathcal{O}(z,\bar{z}) \mapsto \left( -\frac{1}{\bar{z}^2} \right)^h \left( -\frac{1}{z^2} \right)^{\bar{h}} \mathcal{O}\left( \frac{1}{\bar{z}}, \frac{1}{z} \right).$$

Evidently this is still not the result that we want. What happens to the sign in the bracket? Why does no one even bother to mention it (from what I've been able to find)? Was my assumption, that we just need to conjugate the whole expression in the last step false?