Symmetry of real and imaginary parts in FFT
Yes, this is always true if the input to the DFT is real valued. It's called the "conjugate complex symmetry", because
$$
X_{N-n} = {X_n}^*
$$
where $X_n$ is the DFT output and $()^*$ denotes the conjugate. It can be proven by inserting the property into the transformation formula of time domain sequence $x_k$:
$$
X_n = \sum_{k=0}^{N-1}x_ke^{-j\frac{2\pi k n}{N}}\\
X_{N-n} = \sum_{k=0}^{N-1}x_ke^{-j\frac{2\pi k (N-n)}{N}}\\
=\sum_{k=0}^{N-1}x_k e^{-j 2\pi k}e^{j\frac{2\pi k n}{N}}
$$
Using $exp(-j2\pi k) = 1 \:\: \forall \: k$ we find
$$
X_{N-n} = \sum_{k=0}^{N-1}x_k e^{j\frac{2\pi k n}{N}}
$$
Now we exploit that $x_k$ is real for all $k$ to rewrite the above:
$$
X_{N-n} = \sum_{k=0}^{N-1}\left(x_k e^{-j\frac{2\pi k n}{N}}\right)^* = X_n^*
$$
It's now straightforward to show that the same is valid for the IDFT. (real spectrum -> complex conjugate symmetric time domain sequence). Moreover, the opposite statement is also true: if the input to the DFT (IDFT) is complex conjugate symmetric its output is real valued.