Let $B$ be a smooth projective connected variety over $\mathbb C$. Let $\sigma:B\to B$ be an endomorphism of degree one.
Do I understand correctly that $\sigma$ is an automorphism?
I believe this should follow from Zariski's main theorem, I thought it wouldn't hurt to be sure by asking here.
More generally, what kind of assumptions do we really need on $B$ to conclude that every degree one endomorphism is an automorphism?
On
This is true for any normal irreducible algebraic variety $B$ over any field $k$. A (rather easy) proof using Grothendieck groups of varieties over $k$ can be found in § 4, Corollary 8 of this paper.
A few words on the proof. Suppose $f$ is not finite. Then by ZMT, there exists a proper closed subset $F$ of $B$ such that $f^{-1}(B\setminus F)\to B\setminus F$ is an isomorphism, and $f^{-1}(F)\to F$ has connected fibers of positive dimension. In particular, $$\dim f^{-1}(F)>\dim F.$$ Now in the Grothendieck ring $K_0(Var_k)$ of algebraic varieties (= schemes of finite type) over $k$, we have $$[f^{-1}(F)]=[B]-[B\setminus f^{-1}(F)]=[B]-[B\setminus F]=[F].$$ But in the linked paper, it is proved that the dimension of a variety is determined by its class in $K_0(Var_k)$. Contradiction.
Yes, what you want is true, more or less by ZMT as you say. Here's an argument.
First, $\sigma$ is surjective, since it is dominant and $B$ is projective and irreducible.
By ZMT, $\sigma$ has connected fibres, so if it is not an isomorphism, it must contract a curve. But now if $f: X \rightarrow Y$ is a surjective map of smooth proper varieties, with connected fibres, and $f$ contracts a curve, then the Picard number goes down: $\rho(Y) < \rho(X)$. So if $Y$ and $X$ are the same variety, we get a contradiction.
The same argument works as long is $B$ is just $\mathbf{Q}$-factorial (including normal & irreducible) instead of smooth. (Though as QiL'8 points out, this assumption is still much stronger than is needed.)