Let $X$ and $X'$ be two hypersurfaces in $\mathbb{C}^n$ that contain the origin $0$. Consider the following two statements:
The germs of $X$ and $X'$ at $0$ are biholomorphically equivalent, that is they have biholomorphic representatives or equivalently their local rings at $0$ are isomorphic $\mathbb{C}$-algebras.
There exist a biholomorphism $\phi: \Omega\to \Omega'$ of open neighborhoods of $0$ in $\mathbb{C}^n$ such that $\phi(0)=0$ and $\phi (X\cap \Omega) = X'\cap \Omega'$.
Clearly, the second statement implies the first one. Also, the converse is true if they are non-singular at $0$. Is the converse true if they have isolated singularity at $0$?
Answer: Suppose they have isolated singularity. Any isomorphism of germs $f:(X,0)\to (X',0)$ obviously induces a homomorphism of algebras $F:\mathcal{O}_{\mathbb{C}^n,0}\to \mathcal{O}_{\mathbb{C}^n,0}$. Let $\mathfrak{m}$ be the maximal ideal of $\mathcal{O}_{\mathbb{C}^n,0}$. Due to the isolated singularity, the induced map $F:\mathfrak{m}/\mathfrak{m}^2 \to \mathfrak{m}/\mathfrak{m}^2$ is an isomorphism. Then by the inverse function theorem, $F$ is an isomorphism. In particular, from $F$ we can find $\phi: \Omega\to \Omega'$ with the required condition.
