(Inverse?) Fourier transform for finite abelian groups Let $G=\mathbb{Z}_{n_1}\times\cdots\times \mathbb{Z}_{n_m}$ be a finite abelian group and $C(G)$ the algebra of complex-valued functions on $G$ with pointwise-multiplication. Let a basis be given by:
$$\left\{\delta_{(g_1,\dots,g_m)}\mid(g_1,\dots,g_m)\in G\right\}.$$
Let $\mathbb{C}G$ be the group algebra with basis $\{(g_1,\dots,g_m)\mid (g_1,\dots,g_m)\in G\}$, and multiplication the bilinear extension of the group law.
Let $N=\prod_{j=1}^m n_j$.
Consider the following linear map $C(G)\to \mathbb{C}G$:
$$\delta_{(g_1,\dots,g_m)}\mapsto \frac{1}{N}\sum_{(t_1,\dots,t_m)\in G}\left(\prod_{j=1}^m \exp\left(\frac{2\pi i t_jg_j}{n_j}\right)\right)(t_1,\dots,t_m)\qquad \qquad(\star).$$
Let us call this map $T$. Recalling $T(\delta_{(g_1,\dots,g_m)})\in\mathbb{C}G$, some properties I have, if I am not mistaken, are  (the conjugate-linear involution is $g^\ast=g^{-1}$):
$$T(\delta_{(g_1,\dots,g_m)})=T(\delta_{(g_1,\dots,g_m)})^2=T(\delta_{(g_1,\dots,g_m)})^*.$$

Now to say the definition of this map is inspired by Fourier-type ideas is an understatement. The problem is that I have confusion about the correct source and target for the Fourier transform, and the duality between $G$ and its dual is confusing this further. So some references will suggest that $T$ is the Fourier transform while others will suggest it is the inverse... and nowhere have I seen $(\star)$ written down explicitly (when it looks familiar, maybe the $1/N$ is missing). So, some questions:

Is $(\star)$ written down explicitly anywhere?
Is $T$ a Fourier transform? Is is perhaps a Fourier transform $C(G)\to \mathbb{C}G$ but also the inverse of a different Fourier transform
$\mathbb{C}G\to C(G)$?
Is there a similarly explicit expression for the inverse of $T$?

It would be nice to have the correct terminology for this map (and also know that it is invertible).
EDIT: In addition I want $T:C(G)\to \mathbb{C}G$ and it's inverse to be unital in the sense that, where $\mathbf{1}_G$ maps everything one:
$$T(\mathbf{1}_G)=e\qquad\text{ and}\qquad T^{-1}(e)=\mathbf{1}_G.$$
 A: Your $(\star)$ looks very reasonable, but there are many other good choices for it: as you mention, some people normalise by dividing one direction (usually the inverse transform) by $N$, others prefer to put a denominator of $\sqrt{N}$ on both the transform and its inverse. The key choice which is being slid under the rug here is that you have explicitly chosen a set of generators for your group $G$. Someone else might choose a different set of generators and get a different output for the Fourier transform (it would still have all the right properties though). If you'll allow me to delve into some theory for a moment, it may explain the situation.
What's actually going on here is that the Fourier transfom is naturally (meaning there are no choices to be made) a linear map $\mathcal{F} \colon C(G) \to C(\hat{G})$, where $\hat{G} = \operatorname{Hom}_{\mathsf{Grp}}(G, \mathbb{C}^\times)$ is the dual group consisting of all group homomorphisms $G \to \mathbb{C}^\times$, with pointwise multiplication as the group operation. For $G$ finite abelian, we have that $G$ is isomorphic to its dual $\hat{G}$ but not canonically: there are many choices of isomorphisms and none of them is preferred.
An element $\chi \colon G \to \mathbb{C}^\times$ of the dual group is usually called a character. Given a function $f \in C(G)$ and a character $\chi \in \hat{G}$, we can integrate them against each other:
$$ \int_G f \chi = \sum_{g \in G} f(g) \chi(g).$$
The Fourier transform of $f \in C(G)$ is the function $\mathcal{F}f \in C(\hat{G})$ defined by $(\mathcal{F} f)(\chi) = \int_G f \chi$. You can prove that this has all the properties you would expect of something called a Fourier transform (linearity, convolution becomes pointwise product, etc), however it is now a function on the dual group.
There is a natural isomorphism from $G$ to its double dual $\hat{\hat{G}}$, given by $g \mapsto \operatorname{ev}_g$ ("evaluation at $g$"), where $\operatorname{ev}_g(\chi) = \chi(g)$. Using orthogonality of characters it is easily shown that when you take the Fourier transform again, you get $(\mathcal{F} \mathcal{F} f)(\operatorname{ev}_g) = |G| f(g^{-1})$. Therefore the Fourier transform is almost its own inverse: you need to divide by $|G|$ somewhere, and invert or conjugate somewhere. (Compare the usual formulas for the discrete Fourier transform and its inverse: there is a division by $|G|$ and a conjugation).
This is all very well, but often we want to actually do calculations rather than talking about nice coordinate-free formulas, so often we need to choose a particular isomorphism $G \to \hat{G}$. A straightforward way to do this is to do exactly as you have done: pick a set of generators $g_1, \ldots, g_k$ so that $G$ is a product of cyclic groups with orders $n_1, \ldots, n_k$, then define a map $G \to \hat{G}$ by taking the group element $g = m_1 g_1 + \cdots + m_k g_k$ to the character
$$ r_1 g_1 + \cdots + r_k g_k \mapsto \exp\left(\frac{2 \pi i m_1 r_1}{n_1}\right) \cdots \exp\left(\frac{2 \pi i m_k r_k}{n_k}\right)$$
You can see that under this identification, the formula I've given above for the coordinate-free Fourier transform is almost exactly your $(\star)$, the only difference being a division by $|G|$. To take the inverse transform, all you need to do is add/remove the division by $N$, and conjugate the exponentials.
If you would like to read more, there is ample literature out there about dual groups for finite abelian groups. A lovely extension of this is Pontryagin duality which essentially applies Fourier transforms to more groups: for finite cyclic groups you get the DFT, for more general finite groups you get your $(\star)$, for the circle you get the Fourier expansion of a periodic function, and for $\mathbb{R}$ you get the Fourier transform which you might have come across in a differential equations or measure theory course.
