Confusion related to norm of dft
Check the scaling factors used in your DFT/IDFT implementation. There is an arbitrary scaling factor that differs by implementation. Specifically, if you take an input vector $x$ and round-trip it via a DFT followed by an IDFT:

$$
y = IDFT(DFT(x))
$$

Then you may see a couple different results in common implementations:


$y = x$
$y = Nx$, where $N$ is the transform length


Other cases that I wouldn't be too surprised to see:


$y = \sqrt{N}x$
$y = \frac{1}{N}x$
$y = \frac{1}{\sqrt{N}}x$


Which one of these that you see depends on which scaling factors are used in the forward and reverse transforms. If you want the norm of the DFT of $x$ and the norm of $x$ to be the same, then you want the unitary DFT definition:

$$
X[k] = \frac{1}{\sqrt{N}} \sum_{n=0}^{N-1}x[n]e^{\frac{-j2\pi k n}{N}}
$$