Let $X$ be an integral Noetherian scheme that has no automorphisms other than the identity. Is it possible to characterize (in a not completely tautological way) $X$ such that


* there is no finite morphism $X\rightarrow X$ surjective on the underlying topological spaces (other than the identity)?
* there is no finite flat morphism $X\rightarrow X$ surjective on the underlying topological spaces (other than the identity)?


If $X$ is regular, then there is [no distinction](https://mathoverflow.net/a/106575) between two bullet points. 


Among spectra of fields, one can find examples satisfying the first (eq. second) bullet point ($\mathrm{Spec}\,\mathbb{Q}$) or not satisfying the first (eq. second) bullet point ($\mathrm{Spec}\,k$ for $k$ [rigid non-perfect](https://math.stackexchange.com/a/540877)). 



Schemes with no finite morphisms onto themselves
The question as stated seems to me too broad. Nevertheless, in some cases it is actually possible to provide a complete characterization. For instance, ruled surfaces admitting non-trivial, surjective endomorphisms are classified in


N. Nakayama: [*Ruled surfaces with non-trivial surjective endomorphisms*](http://dx.doi.org/10.2206/kyushujm.56.433), Kyushu J. Math. **56**, No. 2, 433-446 (2002). [ZBL1049.14029](https://zbmath.org/?q=an:1049.14029). 


