TITLE: Realizing mapping classes as isometries?
QUESTION [8 upvotes]: Let $\phi : M \to M$ be a diffeomorphism.  Is there a metric $g$ on $M$ and a diffeomorphism $\psi$ isotopic to $\phi$ so that $\psi$ is an isometry with respect to $g$?  I'm guessing the answer is no, since there are manifolds such that all metrics admit no nontrivial isometries and maybe some of these manifolds have a nontrivial mapping class group.
I'm actually particularly interested in the case of the 3-torus which has mapping class group $GL_3(\mathbb{Z})$.  Which of these mapping classes can be realized as isometries for some metric?

REPLY [6 votes]: To complement Dmitri Panov's answer:

Let $\phi$ be a (smooth) diffeomorphism of a closed manifold. Then $\phi$ preserves a Riemannian metric if and only if $\phi$ is isotopic to a diffeomorphism of finite order.

$\Leftarrow$ is clear. Conversely, suppose there's such a metric and let $G$ be its isometry group, which is a compact Lie group. Then the closure $H$ of $\langle\phi\rangle$ in $G$ is a compact abelian Lie group, hence isomorphic to $F\times T$ for some finite group $F$ and some torus $T$. In particular, every connected component of $H$ contains an element of finite order, and this applies in particular to the connected component of $\phi$.

As a corollary, when it holds, the image of $\phi$ in the mapping class group ( = group of diffeomorphisms modulo smooth isotopy, I guess? hence also modulo continuous isotopy) has finite order (this is the contents of Dmitri Panov's answer). I don't know if the converse holds. In other words, I don't know if a mapping class of finite order always has a lift of finite order (comments welcome if you know more!).