I am aware of how Fourier Transformation and Fast Fourier Transformation works, however I do not understand the logic of the inverse of FFT.
Could someone explain why the inverse fourier transformation of $X_k$ is $\dfrac{1}{2^k} X_k$, assuming that $X_k$ is the actual fourier transformation matrix over an Abelian group $(Z_2)^n$
The Fourier transform over $\Bbb{Z}_2^n$ is different from the FFT (which is over $\Bbb{Z}_{2^n}$). The Fourier transform over $\Bbb{Z}_2$ is
$$H' = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$$
so the Fourier transform over $\Bbb{Z}_2^n$ is
$$H'^{\otimes n}$$
Since $H' \cdot H' = 2 I$, it follows that
$$H'^{\otimes n} \cdot \frac{1}{2^n} H'^{\otimes n} = I$$