Is every pseudo-automorphism (self-birational map which does not contract any hypersurface) of a smooth Fano variety of Picard rank $1$ equal to a biregular automorphism?


Remark: For $\mathbb{P}^n$, the answer is yes, and easy: every birational map of degree $>1$ contracts a hypersurface, given by its Jacobian. Same for any projective surface (because birational maps are sequence of blow-ups and blow-downs).



Pseudo-automorphisms on Fano varieties
You don't need your variety, say $X$, to be Fano, only $\mathrm{Pic}(X)=\mathbb{Z}$. A pseudo-automorphism $u$ of $X$ induces an automorphism of $\mathrm{Pic}(X)$, which must be the identity. Let $L$ be a very ample line bundle on $X$; since $u^\*L\cong L$, $u$ induces an automorphism of $H^0(X,L)$ (here you use Hartogs theorem, as Jason pointed out).
Then $u$ induces an automorphism of $\mathbb{P}(H^0(X,L))$ which preserves the image of $X$, hence an automorphism of $X$.


Note that if $K\_X\geq 0$, any birational map is a pseudo-automorphism, and therefore biregular. Of course this doesn't hold for Fano varieties.


