**Question.** Let $V$ be a complex projective manifold of general type (we can even assume that the canonical bundle of $V$ is ample). Suppose $\varphi: V\to V$ is a non-identical automorphism. Can $\varphi$ be isotopic to the identity map (i.e. $\varphi\in Diff\_0(V)$)? 


I hope the answer is no, and this can be easily proven when $K\_V$ is very ample. 


More generally what restrictions are known on smooth manifolds that admit self-diffeos of finite order that are isotopic to identity?



Finite order automorphisms of complex projective manifolds isotopic to identity
The answer to your question is unknown already for surfaces $S$ of general type. 


Note that, if $S$ is simply connected, by a result of Quinn (see "Isotopy of 4-manifolds", Journal of Differential Geometry 1986) every automorphism acting trivially on rational cohomology must be *topologically* isotopic to the identity. 


At any rate, it seems that people conjecture that the answer to your question is **no** for simply connected surfaces of general type. See Catanese's paper "A Superficial Working Guide to Deformations and Moduli" (arXiv:1106.1368), Section 1.4 for further details. In this paper, complex manifolds which do not admit non-trivial automorphisms isotopic to the identity are called *rigidified*. 


