If we know that some homomorphism $\phi: GL_{2}(\mathbb{R})\to G$ has a non-trivial kernel, consisting of invertible square real matrices, and we know that these matrices are similar to diagonal matrices by conjugation, then for some matrix $B \in GL_{2}(\mathbb{R})$ there exists some matrix $P$, such that $PBP^{-1} = A$, and thus $A$ is also in the kernel of $\phi$. However, by Proposition 7 on page 82 in Abstract Algebra by Dummit and Foote, we can read that "A subgroup $N$ of the group $G$ is normal if and only if it is the kernel of some homomorphism". But if $N$ is normal in our case, then this implies that $gNg^{-1} \subset N$ for all $g \in G$.
But does that imply that $PBP^{-1} = A$ for any $P \in GL_{2}(\mathbb{R})$. Am I mistaken or is this really possible? From my Linear Algebra experience, I thought there is a unique matrix $P$ for every matrix $B$, such that $PBP^{-1} = A$. Please help me clarify this interesting moment.
First, not all matrices can be diagonalized. So given a random $B \in \mathrm{GL}_2(\mathbb{R})$, we could have that $B$ is diagonalizable, but only over the complex numbers (this occurs when $B$ has complex eigenvalues). Or even worse, $B$ might not be diagonalizable over any field of scalars (it has a non-trivial Jordan block).
Next, given $PBP^{-1}=A$, yes, if $B$ is in the kernel, then $PBP^{-1}=A$ must be in the kernel as well (kernels are normal subgroups and normal subgroups are closed under conjugation). However, there's no reason to assume that $QBQ^{-1}=C$ is the same matrix (if we swap out $P$ for some other invertible matrix $Q$). However, $C=QBQ^{-1}$ would still be in the kernel (it just may not be the same element of the kernel).
Finally, IF a matrix $B$ is diagonalizable, then any matrix whose columns form a basis of eigenvectors, $P$, will diagonalize $B$.
Take the first column of $P$ (some eigenvector of $B$), rescale it (just not by zero) and you'll still have an eigenvector of $B$. Rescaling a vector won't change linear independence, so $Q$ (where $Q$ is $P$ with its first column rescaled) still diagonalizes $B$. Therefore, there are infinitely many matrices that diagonalize $B$!