Let $X$ be a variety with at most canonical singularities and ample canonical divisor $K_X$. Then $\operatorname{Aut}(X)=\operatorname{Bir}(X)$.
The canonical ring $R$ of $X$ is finitely generated. Furthermore $$X\cong\operatorname{Proj}(R)=X_\text{can}$$ since $K_X$ is ample and $X$ has at most canonical singularities. Now, any birational automorphism $f:X\rightarrow X$ induces an automorphism of the canonical ring $R$ which in turns induces a biregular automorphism of $X$. So $f$ itself is biregular.
More generally, if $X,Y$ are projective varieties with $K_X$, $K_Y$ ample and at most canonical singularities, then any birational map $f:X\rightarrow Y$ is indeed biregular.
