From the condition $[A,B]=A$, what can I say about $B$? I'm struggling in understanding the meaning of this condition that I found in an operator equation:
\begin{equation}
[A,B]=A
\end{equation}
where both $A$ and $B$ are hermitian operators. What can I say about the operator $B$?
 A: If A and B are hermitian, then you need, instead,
$$
[A,B]= i A ,
$$
for consistency under taking the adjoint. So A is neither hermitian nor antihermitian, even though, conventionally, B is taken to be hermitian in applications. Let's stick with $[A,B]=  A$.
These are the prototypes for the shift and counter operators, used routinely in quantum mechanics. 
Specifically, for an eigenvector v  of B with real eigenvalue λ,
$$
B v = \lambda v ,
$$
utilize your simplest nonabelian Lie algebra you specified, to observe
$$
BAv = A(B-1)v, 
$$ 
and hence 
$$
B(Av)=(\lambda-1)(Av).
$$
That is, A defines new eigenvectors with down-shifted eigenvalues of B, it "ladders" the state v down.
A simple realization of the algebra is $A=x$ and $B=-x\partial_x$. Consider what A does to the eigenvectors $x^{-\lambda}$ of B.
Formally, this is the simplest nonabelian Lie algebra there is, with just one commutator, that of the affine group in one dimension . The upper-triangular 2x2 matrix representation thereof is given in the WP link provided.
You probably do not need to dwell on its delightful group structure. In QM, people use number operators for B and lowering operators a for A.  
