Realizing mapping classes as isometries? 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?
 A: For any compact manifold you need your diffeo to be of finite order in the mapping class group. Here is a proof.
Proof. Suppose $M$ is compact and $\varphi$ is an isometry of $M$ for some metric $g$. Let's prove that $\varphi^n$ is isotopic to identity for some $n$. Fix a point $x\in M$ and an orthonormal basis  $(e_1,\ldots,e_n)$ at $x$. Let $E$ be the fibration over $M$ consisting of bases of orthonormal vectors in $TM$. Let $\tilde \varphi$ be the lift of $\varphi$ to $E$. Then there exits a sequence $n_k$ such that $\tilde \varphi^{n_k}(x,e)$ tends to $(x,e)$ (this is easy). It is easy to see  that for $k$ large enough such $\varphi^{n_k}$ is isotopic to identity (since it is an isometry and moves points on $M$ just a bit).
A: 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!).
A: As you suspect, in general, no.
For example, if $M$ is compact, and $\psi:M\to M$ fixes a metric $g$ on $M$, then the closure of $\{\psi^k\ |\ k\in\mathbb{Z}\ \}$ is a compact abelian subgroup of $\mathrm{Isom}(M,g)$, and, hence, its identity component is a torus, so $\psi$ must be isotopic in $\mathrm{Isom}(M,g)$ to a $g$-isometry of finite order.  In particular, the action of $\psi$ on the deRham cohomology of $M$ must have finite order, as must any diffeomorphism $\phi:M\to M$ homotopic to $\psi$.
Conversely if the closure of $\{\psi^k | k\in\mathbb{Z}\ \}$ in the compact open topology is a compact group, then, yes $\psi$ fixes a metric $g$.
