"Centralizer" of an irreducible character? There is a chance that my formulations of the questions is wrong (in which case I apologize), but I want to give it a try as I thought these questions would be natural to ask, and there should have been answers already.
Let $G$ be a finite group, and $\mathbb{C}$ the field of complex numbers. So the group algebra $\mathbb{C}G$ can be decomposed as full matrix algebras:
$\mathbb{C}G = B_1 \oplus B_2 \oplus \dots \oplus B_k$
where $k$ is the number of conjugacy classes of $G$, and each $B_i$ is isomorphic to the full $d_i \times d_i$ matrix algebra $M_{d_i}(\mathbb{C})$. Each $B_i$ is a subalgebra, or 2-sided ideal of $\mathbb{C}G$.
Consider the regular representation of $G$ by right multiplication. So each $B_i$ can be decomposed as direct sum of $d_i$ copies of isomorphic irreducible $\mathbb{C}G$-modules. This decomposition is only unique up to isomorphism, but not unique. Let's say one of the decomposition is as follows:
$B_i = V_{i,1} \oplus V_{i,2} \oplus \dots \oplus V_{i,d_i}$
Each of the above summands is a $d_i$-dimensional irreducible (right) $\mathbb{C}G$-module affording the same irreducible character $\psi$ of $G$. Now here are the questions:
As $B_i$ is also a left $\mathbb{C}G$-module, if we fix one of the summands, say $V=V_{i,1}$, and consider the set $N_G(V)$ of elements in $G$ which fixes $V$ from the left, i.e. $N_G(V) = \{g \in G | gV = V\}$, I think this is a subgroup of $G$? For instance if $d_i = 1$ then $N_G(V) = G$. But if $d_i > 1$ it's definitely not $G$. If I'm correct (that $N_G(V)$ is a subgroup of $G$), is there any characterization of this subgroup of $G$? Obviously this subgroup depends on the choice of $V$, but are they all isomorphic or conjugate in $G$ for all the irreducible (right) $\mathbb{C}G$-submodules isomorphic to $V$? What is the relationship between these subgroups and $\psi$?
Again I feel like if this reasoning were correct, there should have been discussions about this topic in the literature already (I'm not aware of any), so it's quite possible I made a mistake somewhere. I hope experts can point me to the right direction.
 A: Lets define the ring $R:=End(V)$, then we have that $R$ is a summand of $k[G]$, and since the actions are both left and right multiplication, we can phrase this problem as, for a fixed irrep $V$ of $G$, and a minimal right $R$ submodule $V\cong W\subset R$, describe the stabiliser of $W$ by $R$ on the left, then describe the preimage of this in $G$ under $\rho:G\rightarrow R^*$.
This first question is entirely about matrix algebras, so lets view it as such. Then for a matrix $M$, its not too hard to check that the right ideal $M\cdot R$ will be minimal if and only if $M$ is rank $1$, and the ideal will be all matrices with column span contained in that of $M$.
We can also check that the action of $R^*$ acts transitively on the set of these minimal right submodules by left multiplication, from the description of rank one matrices as outer products of vectors. So all the stabiliser algebras of these are conjugate in $R$, so to identify this subalgebra, we can look at the module generated by $e_{1,1}$, which is those matrices with support in the top row only. Then the stabiliser is the "parabolic" of the form $M_{1\times 1}\oplus M_{n-1\times n-1}$ in block diagonal form.
So we can recognise these subgroups in $G$ as the preimages of conjugates of these subalgebras in $R$, in particular, there is very little canonicity in these subgroups, its very dependent on our choice of decomposition.
For instance for a fixed irrep, and a subgroup $H$ which acts by scalars on a one dimensional subspace $U:=k\cdot v$ inside $V$, then the orthogonal (canonical inner product is floating around) projection onto $U$ will have right ideal minimal, and $H$ will be in its stabiliser.
Edit for more info:
To answer your question what information do these stabilisers give us, taking for any $v\in V$ the stabiliser of the line containing $v$, then these subgroups are precisely the subgroups for $H$ which $V$ occurs in an induced linear representation from $H$, so they inform you about how to express your representation as a sum of reps induced from linear chars of subgroups. Taking this to its conclusion, one gets an "explicit", canonical description of $\chi_V$ as an alternating sum over induced linear characters from subgroups by stratifying the space $\mathbb{P}V$ by its stabilisers, with each induced character (nearly) weighted by the compactly supported Euler characteristic of the stratum according to it.
This circle of ideas is explicit Brauer induction, and can be read about in Snaiths book of that same name.
