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.

This is also true for every smooth Fano variety $X$, with any Picard number. One can see it using Mori dream spaces: $X$ is a Mori dream space (by BCHM), and hence has (up to isomorphisms) only finitely many "small $\mathbb{Q}$-factorial modifications" (SQM) = $f\colon X$-->$Y$ birational, isomorphism in codimension one, with $Y$ projective, normal, and $\mathbb{Q}$-factorial. These SQMs correspond to the chambers in the decomposition of the cone of movable divisors in $\mathcal{N}^1(X)$, the chamber corresponding to $f$ being $f^*\text{Nef}(Y)$. For arbitrary Mori dream spaces, it can happen that some $Y$ is isomorphic to $X$, and then $X$ has a pseudo-automorphism. But if $X$ is Fano this is impossible, because the anticanonical class is in the interior of the chamber $\text{Nef}(X)$, so it will not be ample on any other model $Y$.

It seems to me that the Fano variety $X$ does not need to be smooth. It is enough to have that $-K_X$ is $\mathbb{Q}$-Cartier. Indeed, in this case we can embed $X$ in a projective space $\mathbb{P}^n$ with the linear system $|-mK_X|$ for $m\gg 0$.

For any pseudo-automorphism $\phi:X\dashrightarrow X$ we have that $\phi^{*}(-mK_X) = -mK_X$, and hence $\phi$ induces an automorphism of $\mathbb{P}^n$ stabilizing $X\subset\mathbb{P}^n$.