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.