Let $X$ be a projective complex manifold. Under what condition do we have the equality $\operatorname{Aut}(X)=\operatorname{Bir}(X)$? Here $\operatorname{Aut}(X)$ denotes the group of holomorphic automorphisms of $X$ and $\operatorname{Bir}(X)$ the group of birational morphisms of $X$.
I am interested in the case when $\dim_{\mathbb{C}}X=2,3$. Maybe there are not universal criteria, so I would appreciate your providing me with any examples for which the equality holds.
To complete the answer of Diverietti and the comment of Roy Smith, here is a statement which might interest you:
Theorem If $X$, $Y$ are varieties over a field $k$, assume $X$ is smooth and $Y$ proper containing no rational curves. Then any rational map $X\dashrightarrow Y$ is everywhere defined.
You can find that statement in Debarre's book Higher-Dimensional Algebraic Geometry, Corollary 1.44 p. 31.
In particular, if $X$ is smooth projective and contains no rational curves, then its automorphism group is equal to the group of its birational endomorphisms.
It also holds for minimal surfaces of Kodaira dimension $\kappa\geq0$.
I realize that no one addressed the question of the OP about Calabi–Yau manifolds. Two remarks:
The equality $\mathrm{Bir}(X)=\mathrm{Aut}(X)$ holds if $K_X$ is nef and $\mathrm{Pic}(X)=\mathbb{Z}$ (see for instance the introduction of Chen - Rational self maps of Calabi–Yau manifolds). This applies in particular to Calabi–Yau complete intersections (of dimension $\geq 3$).
There are examples of birational automorphisms of holomorphic symplectic manifolds which are not biregular, see §6 of Beauville - Some remarks on Kähler manifolds with $c_1 = 0$.
A simplest example is a curve $C$, then $\operatorname{Bir}(C)\cong\operatorname{Aut}(C)$.
A careful note that $\operatorname{Bir}(X)$ is not a group scheme in general. Moreover if $X$ and $Y$ are birational then $\operatorname{Bir}(X)$ is not isometric with $\operatorname{Bir}(Y)$
So we need to add some condition on $X$ in the sense of Minimal Model Program such that in scheme group theoretic sense $\operatorname{Bir}(X)$ behaves well since we have lack of well defined multiplication rule. In fact if $X$ is a minimal model with terminal singularities it is a theorem that $\operatorname{Bir}(X)$ is a group scheme as an Abelian variety. In general $$\operatorname{Aut}^o(X)\cong\operatorname{Bir}^o(X)$$
where $\operatorname{Aut}^o$ is the identity connected component. See Hanamura - On the birational automorphism groups of algebraic varieties.
Let $X$ be a projective variety. Then we have the following isomorphism from MMP:
$$\operatorname{Aut}(X_\text{min})\cong\operatorname{Bir}(X_\text{min})$$ where $X_\text{min}$ is the minimal model of the projective variety $X$.
For any surface of non-negative Kodaira dimension, we have $$\operatorname{Aut}(X)\subset\operatorname{Bir}(X)\cong\operatorname{Bir}(X_\text{min}) \cong\operatorname{Aut}(X_\text{min}).$$
If $X$ is a Fano variety of $\dim\geq 5$ and its Picard group is generated by anticanonical divisor of the variety $X$, then it is conjectured (see Chel'tsov - Birationally rigid Fano varieties) that $\operatorname{Aut}(X)\cong\operatorname{Bir}(X)$.
Moreover for moveable log pair $(X,M_X)$ which is log Calabi–Yau pair, with at worst canonical singularities then $\operatorname{Aut}(X)\cong\operatorname{Bir}(X)$.
One example: This holds for abelian varieties, because a rational map to an abelian variety is always regular.
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.
A large class of compact complex manifolds for which (more generally) $$ \operatorname{Aut}(X)=\operatorname{Bim}(X) $$ holds is given by Kobayashi hyperbolic compact complex spaces. Here $\operatorname{Bim}(X)$ is the group of bimeromorphic automorphism.
A compact complex space $X$ is Kobayashi hyperbolic iff there is no non-constant holomorphic map $f\colon\mathbb C\to X$. For instance, by Liouville's theorem, a compact complex space $X$ is hyperbolic as soon as its universal cover is a bounded domain in $\mathbb C^n$. Other examples are given by compact complex manifolds whose cotangent bundle is Griffiths positive (or, more generally, with ample cotangent bundle).
If $X$ is moreover projective, it is conjectured by Lang that being hyperbolic should be equivalent to have only subvarieties of general type.
This latter class of projective manifolds (of general type, with all subvarieties of general type) have indeed the property your are asking for, too. This is because the indeterminacy locus of a birational map is covered by rational curves (and cannot be of general type, nor hyperbolic).
