I want to build some intuition around the important Great Orthogonality theorem.
The Great Orthogonality theorem states that for a finite group $G$, we have (in a particular form of the theorem): $$\frac{1}{N(G)}\sum_{g\in G}\left(D^\dagger(g)\right)_{ij} \left(D(g)\right)_{jk} =\delta_{ik}$$ where $N(G)$ is the number of elements of a finite group $G$ and $D$ is an irreducible representation of $G$. A word on notation: in the above, the subscripts denote the entries of a matrix. For example, for a matrix $A$, $A_{ij}$ is the element of $A$ at the $i$-th row and $j$-th column. Hence, the above relation could be written as $\frac{1}{N(G)}\sum_{g\in G}D^\dagger(g)D(g)=id_{D(g)}$, where $id_{D(g)}$ is the identity element in the representation $D$ of $G$.
In order to gain some intuition about this, I think as follows: since we know from the unitarity theorem that any representation of a finite group is equivalent to a unitary representation, using the unitary representation of the above representation, $D(g)$, the above statement is satisfied immediately. So, in this sense, the above form of the theorem gives a natural generalization of the case where we use a unitary representation to the case of using $\it{any}$ representation. Is my understanding correct and precise?
Moreover, in the above form of the theorem, it roughly seems like "averaging out" the action of the representation $D(g)$ (I am not talking about the unitary representation now) on whatever space it acts on and that "averaged" action leaves anything it acts on invariant (does not change it). I cannot really precisely understand this; I can only roughly understand it by saying that since $G$ is a group, $\exists\ g^{-1}\in G$ such that $g\cdot g^{-1}=id_G,\ \forall g\in G$ and knowing that $D(g^{-1})=D^{-1}(g)$ (where in the unitary representation $D^{-1}(g)=D^\dagger(g)$); using these facts I can get a rough idea of how the averaged action of the representation of the group does nothing, because for every action of the representation of $g$ there is also the "opposite" action of the rep of $g^{-1}$, something analogous to averaging out all the unit vectors in $R^n$, where for every unit vector we have a unit vector of opposite direction (although here we do no talk about the action of something on something). Is the intuitive picture presented in this paragraph sufficient or is there something missing?
Last, the theorem for different (inequivalent) representation states that $$\frac{1}{N(G)}\sum_{g\in G}\left(D^{\dagger}(g)\right)_{ij} \left(R(g)\right)_{jk} =0$$ with $D$ and $R$ being different and inequivalent irreducible reps of $G$. The way that I understand this is that the action of an irreducible rep is independent of the action of that of an inequivalent irreducible rep, in a certain sense. It is like the analog of an orthogonality relation between vectors. But, this breaks down because the above does not hold when we don't sum over the elements of $G$, so I don't really have a good idea on why it the above holds.