Conjugate symmetry of the DFT of real-valued sequences
We can! In your example, $N=4$, and the DFT is real-valued, so you get $X[k]=X[4-k]$, and that's true:
$$\begin{align}k=0:\;X[0]&amp;=X[4]=0\\k=1:\;X[1]&amp;=X[3]=0\\k=2:\;X[2]&amp;=X[2]=-4\end{align}$$
Note that by definition the DFT is $N$-periodic, so $X[N]=X[0]$.