What is the difference between $D^{(s_{+},s_{-})}(Lambda)$ and $D^{(s_{-},s_{+})}(Lambda)$?

I’m studying QFT and confused about the difference between $D^{(s_{+},s_{-})}(Lambda)$ and $D^{(s_{-},s_{+})}(Lambda)$.

For example, when we make the vector representation, we don’t take $D^{(frac{3}{2},0)}(Lambda)$ but $D^{(frac{1}{2},frac{1}{2})}(Lambda)$. This is because (1) $D^{(frac{3}{2},0)}(Lambda)$ is $4 rightarrow 4$, and (2) $D^{(frac{3}{2},0)}(Lambda)$ become $D^ {(0,frac{3}{2})}(Lambda) $ under the parity inversion. I understand (1), but I can’t understand what is the matter of (2). Both $D^{(frac{3}{2},0)}(Lambda)$ and $D^ {(0,frac{3}{2})}(Lambda) $ are decomposed to $D^{(frac{3}{2})}(Lambda)$, so I feel they are almost same.

I know $s_{pm}$ is a spin number of $J_{pm}=frac{1}{2} (Lpm iM)$, so $D^{(frac{3}{2},0)}(Lambda)$ means $J_{+}^{z}rightarrow3/2, J_{-}^{z}rightarrow 0 $ and $D^ {(0,frac{3}{2})}(Lambda) $ means $J_{+}^{z}rightarrow 0, J_{-}^{z}rightarrow 3/2$, but I cannot understand why is $D^ {(0,frac{3}{2})}(Lambda) $ not equivalent to $D^{(frac{3}{2},0)}(Lambda)$.

Can anyone help me?

References:

1. Zee’ s Group theory in a nutshell, p.458.

2. Weinberg’ s QFT section 5.7 of volume 1.