I'll share here two proofs of the fact I am interested in under the assumption that the origin is the only singular point.

Proof 1: (This is equivalent I think to Torsten's answer, and was a explained to me by a colleaguecolleagues - thank you David and Mike) Let $F: V(I) \rightarrow V(J)$ be an isomorphism. Then since $0$ is the only singular point in $V(I)$, then it is mapped to $0 \in V(J)$. Now the derivative of $F$ at $0$, call it $DF$, is a linear map in $\mathbb{C}^n$ that takes the tangent cone of $V(I)$ at $0 \in V(I)$ to the tangent cone of $V(J)$ at $0 \in V(J)$. But these tangent cones are $V(I)$ and $V(J)$, respectively. Thus, $DF$ is the required linear map taking $V(I)$ onto $V(J)$.

Proof 2: There are some technical details missing here. Let $F: V(I) \rightarrow V(J)$ be an isomorphism. Again, $F$ must take $0$ to $0$, under the assumption that $0$ is the only singular point. Thus, $F$ has the form $$F(z) = Az + \textrm{ higher order terms .}$$ Now define $F_t$ by $$F_t(z) = tF(z/t) .$$ Since $I$ and $J$ are homogeneous, $V(I)$ and $V(J)$ are invariant under scalings, so $F_t$ is again an isomorphismof $V(I)$ and $V(J)$. But $F_t$ has the form $$F_t(z) = Az + \frac{1}{t}(\textrm{higher order terms}).$$ Taking $t \rightarrow \infty$, we converge to the isomorphism (hopefully) $z \mapsto Az$.

That's for the case when $0$ is the only singular point. In fact, all that is used is that there is an isomorphism taking $0$ to $0$. Is it true that whenever there is an isomorphism between $V(I)$ and $V(J)$, there is also isomorphism that fixes $0$? (here $I$ and $J$ are assumed homogeneous, of course).

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I'll share here two proofs of the fact I am interested in under the assumption that the origin is the only singular point.

Proof 1: (This is equivalent I think to Torsten's answer, and was a explained to me by a colleague) Let $F: V(I) \rightarrow V(J)$ be an isomorphism. Then since $0$ is the only singular point in $V(I)$, then it is mapped to $0 \in V(J)$. Now the derivative of $F$ at $0$, call it $DF$, is a linear map in $\mathbb{C}^n$ that takes the tangent cone of $V(I)$ at $0 \in V(I)$ to the tangent cone of $V(J)$ at $0 \in V(J)$. But these tangent cones are $V(I)$ and $V(J)$, respectively. Thus, $DF$ is the required linear map taking $V(I)$ onto $V(J)$.

Proof 2: There are some technical details missing here. Let $F: V(I) \rightarrow V(J)$ be an isomorphism. Again, $F$ must take $0$ to $0$, under the assumption that $0$ is the only singular point. Thus, $F$ has the form $$F(z) = Az + \textrm{ higher order terms .}$$ Now define $F_t$ by $$F_t(z) = tF(z/t) .$$ Since $I$ and $J$ are homogeneous, $V(I)$ and $V(J)$ are invariant under scalings, so $F_t$ is again an isomorphismof $V(I)$ and $V(J)$. But $F_t$ has the form $$F_t(z) = Az + \frac{1}{t}(\textrm{higher order terms}).$$ Taking $t \rightarrow \infty$, we converge to the isomorphism (hopefully) $z \mapsto Az$.

That's for the case when $0$ is the only singular point. In fact, all that is used is that there is an isomorphism taking $0$ to $0$. Is it true that whenever there is an isomorphism between $V(I)$ and $V(J)$, there is also isomorphism that fixes $0$? (here $I$ and $J$ are assumed homogeneous, of course).