Fourier transform of a Fourier transform
Applying the discrete Fourier transform (DFT) twice to a time-domain signal, should produce the original time-domain signal in a circularly time-reversed, and linearly scaled (by N) form; as the inverse DFT is very similar to the forward DFT, except the sign of the complex exponential,a linear scale by 1/N.
The inverse DFT of $X[k]$ is:
$$ x[n] = \frac {1}{N} \sum_{k=0}^{N-1} {X[k]e^{j \frac{2\pi}{N} k n}}$$
By reverting the sign of the exponent $e^{j \frac{2\pi}{N} k n}$ into negative, by a change in the sign of the $n$ of $x[n]$ what you get is:
$$  N x[(-n)_N] = \sum_{k=0}^{N-1} {X[k]e^{-j \frac{2\pi}{N} k n}} $$
which is N times the circularly reversed (obtained via modulus operator denoted by $(.)_N$) version of the original signal $x[n]$, which is obtained by an apparently forward but effectively inverse DFT applied to $X[k]$ which is treated as a time-domain sequence, which is what you do at your second stage of applying forward FFT to $X[k]$ of $x[n]$.