Let $n\geq 2$ and $\mathbb{P}^n(\mathbf{C})$ be the complexe projective space of dimension $n$. Let $H\subseteq \mathbb{P}^n(\mathbf{C})$ be a hypersurface of degree $d$ where the coordinates in $\mathbb{P}^n(\mathbf{C})$ are chosen to be $x=[x_1,\ldots,x_{n+1}]$. Since $H$ is a hypersurface of degree $d$, it is defined by the zero locus of a homogeneous polynomial $F=F(x_1,\ldots,x_{n+1})$ of degree $d$. Similarly, let $H'\subseteq \mathbb{P}^n(\mathbf{C})$ be a second hypersurface of degree $d$ where the coordinates in $\mathbb{P}^n(\mathbf{C})$ are chosen to be $x'=[x_1',\ldots,x_{n+1}']$ and the defining homogeneous polynomial of $H'$ is $F'=F'(x_1',\ldots,x_{n+1}')$.

Let $\phi:H\rightarrow H'$ be a biregular map. So $\phi=[\phi_1,\ldots,\phi_{n+1}]$ where the $\phi_i$'s may be assumed to be homogeneous polynomials of degree $e\geq 1$ such that $\gcd(\phi_1,\ldots,\phi_{n+1})=1$. By convention the trivial polynomial $0$ is consider to be homogeneous in any degree.

**Question:** Do we necessarily have $e=1$ ?

Let me give an interesting special case where the question has a positive answer.

**Proposition:** Assume that $P_0=[0,0,\ldots,0,1]$ is a point, both on $H$ and $H'$, such that $\phi(P_0)=P_0$. Assume furthermore that if $x'\in H'$ is such that $x_1'=0$ then necessarily $x_2'=0$. Then $e=1$.

**Proof** Let us suppose that $e\geq 2$. We will reach a contradiction by showing that the variety set $\phi^{-1}(P_0)$ has a too large total multiplicity. Notice that $\phi^{-1}(P_0)$ correspond to the zero locus of the following set of polynomials in the variables $x_1,\ldots,x_{n+1}$:
$$
F=0, \phi_1=0,\phi_3=0,\ldots,\phi_{n+1}=0.
$$
This variety, say $V\subseteq H$, must be zero dimensional becaue $\phi$ is a finite to one map. Therefore, by Bezout theorem, this intersection set, counting multiplicities, has size $de^{n-1}$. Since $\phi$ is injective and $\phi(P_0)=P_0$, it follows that this set consists of a single point, namely $P_0$, which has multiplicity $de^{n-1}$. On the other hand, the subvariety $V\subseteq H$ may also be characterized as being the zero locus of
$$
F=0,x_1=0,x_2=0,\ldots, x_{n}=0.
$$
Again by Bezout theorem, the multiplicity of $P_0$ must be $d$. But $d<de^{n-1}$. Contradiction.

Note that the result of the previous proposition covers the case of an elliptic curve written in Weierstrass normal form.