For a good overall introduction click on Discrete Fourier Transform. Here we'll try to approach the subject in terms of operators commuting with permutation groups because this provides a very natural link to linear algebra and, amazingly enough, is open to considerable generalization.
The role of translation is played by permutation matrices. On ${\mathbb C}^4$, for instance, this permutation matrix is $$P \ = \ \left[\begin{array}{cc} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right]\,.$$ One way of exhibiting translation is to note that on each ${\bf v}$ in ${\mathbb C}^4$, $$P\,{\bf v} \ = \ \left[\begin{array}{cc} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right] \left[\begin{array}{cc} v_1\\ v_2\\ v_3\\ v_4 \end{array}\right] \ = \ \left[\begin{array}{cc} v_4\\ v_1\\ v_2\\ v_3 \end{array}\right]\,.$$ But what's going to play the role of modulation? That requires bringing in a notion of periodicity.Now let's try to factor the Fourier transform matrix because this is one very natural way of seeing why the Fast Fourier Transform works. The idea is to represent ${\cal F}$ as a product of matrix operations in which the rows of each matrix contain only 2 non-zero elements, thereby reducing number of calculations despite multiplying more matrices. Again for simplicity we'll look first at the $n = 4$ case so that $w = e^{2\pi i \over 4} = i = \sqrt{-1}$. Then as calculations show,
$${\cal F}_4\ = \ \frac{1}{2}\left[\begin{array}{cc}1&1&1&1\\ 1&i&-1&-i\\ 1&-1&1&-1\\ 1&-i&-1& i\end{array}\right]\ = \ \left[\begin{array}{cc}1&1&0&0\\ 0&0&1&i\\ 1&-1&0&0\\ 0&0&1&-i\end{array}\right] \left[\begin{array}{cc}1&0&1&0\\ 0&1&0&1\\ 1&0&-1&0\\ 0&1&0&-1\end{array}\right]$$When $n = 8$ then $w = e^{2\pi i / 8} = e^{\pi i / 4}$ so that $$w^2\,=\,i,\ \ \ w^3\, =\, iw, \ \ \ w^4\, =\, -1, \ \ \ w^5\,=\,-iw,\ \ \ w^6\,=\,-i,\ \ \ w^7\,=\, -iw.$$ In this case $${\cal F}_8 \ = \ \frac{1}{\sqrt{8}}\left[\begin{array}{cc}1&1&1&1&1&1&1&1\\ 1&w&i&iw&-1&-w&-i&-iw\\ 1&i&-1&-i&1&i&-1&-i\\ 1&iw&-i&w&-1&-iw&i&-w\\ 1&-1&1&-1&1&-1&1&-1\\ 1&-w&i&-iw&-1&w&-i&iw\\ 1&-i&-1&i&1&-i&-1&i\\ 1&-iw&-i&-w&-1&iw&i&w\end{array}\right]$$ Now it's far from obvious just what the corresponding matrix product is, but at least a calculation shows that $${\cal F}_8 \ = \ \frac{1}{\sqrt{8}}\left[\begin{array}{cc}1&1&1&1&1&1&1&1\\ 1&w&i&iw&-1&-w&-i&-iw\\ 1&i&-1&-i&1&i&-1&-i\\ 1&iw&-i&w&-1&-iw&i&-w\\ 1&-1&1&-1&1&-1&1&-1\\ 1&-w&i&-iw&-1&w&-i&iw\\ 1&-i&-1&i&1&-i&-1&i\\ 1&-iw&-i&-w&-1&iw&i&w\end{array}\right]$$