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. 



Biregular maps between hypersurfaces of the same degree
Since $\phi$ is biregular, it induces an isomorphism $\mathrm{Pic}(H')\stackrel{\sim}{\rightarrow }\mathrm{Pic}(H)$, which maps $\mathcal{O}\_{H'}(1)$ to $\mathcal{O}\_H(e)$. If $e>1$ this implies that $\mathcal{O}\_{H'}(1)$ is divisible in $\mathrm{Pic}(H')$, which is impossible if your hypersurfaces have reasonable singularities, in particular if they are smooth.


