Mapping Class Group of Simply Connected Spaces

Question

I was wondering the following: If we take $M$ to be some orientable, simply-connected $n$-manifold. What can be said about $\pi_0(Homeo(M))$? We know that $\pi_1(M)=0$ and I know that the group is divided into orientation-preserving and orientation-reversing homeomorphisms which will not be isotopic. It seems reasonable to believe that many of the elements of $Homeo(M)$ are isotopic (and my reasoning is based on intuition involving simply connected spaces so I may be (probably) severely wrong). Any insight is much appreciated, thank you.

Answer 1

Here's a recent paper with the answer for 4-manifolds: an automorphism of $M$ induces an automorphism of of the homology of $M$ which preserves the intersection form or multiplies it by $-1$ according as orientation is preserved or reversed. On the orientation-preserving component this is an isomorphism from the mapping class group to the intersection-equivariant automorphisms of $H_2(M)$.

These groups can be rather big: for instance the E8 manifold, constructed by Freedman on the way to the classification of simply connected 4-manifolds, has an orientation-preserving mapping class group of order $696729600.$