Elements of finite order in mapping class groups of high dimensional manifolds Let $M$ be a manifold with boundary.  Consider the following groups:
(1) $\pi_0(\operatorname{Diff}(M,\partial M))$.
(2) $\pi_0(\operatorname{Homeo}(M,\partial M))$.
(3) $\pi_0(\operatorname{HomEq}(M,\partial M))$.
That is, isotopy (resp. isotopy and homotopy) classes of diffeomorphisms (resp. homeomorphisms and homotopy equivalences) relative to $\partial M$.

I would like to find conditions on $M$ which guarantee that one of the groups (1)/(2)/(3) has no element of finite order.

My motivation for this is to generalize from the case $\dim M=2$, when as long as $\partial M\ne\varnothing$, none of the groups (1)/(2)/(3) has an element of finite order (This is left as an exercise for the reader.  Hint: an element of finite order in $\operatorname{MCG}(\Sigma_g)$ fixes some hyperbolic structure).
Of course, if $\dim M=2$, then (1)=(2)=(3).  This question is really about finding a proper generalization of the result for $\dim M=2$ to higher dimensions, so I'm leaving it open as to which of (1)/(2)/(3) this question is really about.  We remark that (1) seems unlikely to be the right group to consider; for instance when $(M,\partial M)=(D^n,S^{n-1})$ and $n>4$, then it is the group of exotic spheres in dimension $n+1$.
 A: In the case of simply connected 4-manifolds, a famous theorem of Michael Friedman asserts that $\pi_0 Homeo(M)$ is isomorphic to the group of automorphisms $Aut(Q)$ of the intersection quadratic form $Q$ on the middle homology.  This is an arithmetic group, and hence it contains a finite index subgroup that is torsion free.  However, $Aut(Q)$ will almost always have torsion, except possibly in some low-rank cases.
One way of thinking about this is:  For surfaces, the map from MCG to $Aut(Q)=Sp_{2g}(\mathbb{Z})$ (sending a homeomorphism to the induced automorphism of homology) has a huge kernel, namely the Torelli group.  But in dimension 4 under the simply connected hypothesis this map is an isomorphism.  
A: Here's one case where you can answer (3). 
Let $M$ be a compact hyperbolic manifold of dimension $ \geq 3$.  By Mostow rigidity, $HomEq(M)$ has the same homotopy-type as its isometry group, a finite group.  By design, $Isom(M) \simeq Out(\pi_1 M)$.   This latter isomorphism is because $M$ is a $K(\pi,1)$ space. 
So $\pi_0 HomEq(M)$ contains elements of finite order if and only if $Out(\pi_1 M)$ does, if and only if this hyperbolic manifold has symmetries. 
