What are some examples of knots $K\subset S^3$ such that the mapping class group of $S^3\_{1/n}(K)$ is trivial? I guess for hyperbolic knots with no symmetry in the complements are good candidate as one might argue that maybe for sufficiently large $n$, the resultant manifold have trivial mapping class group. But can some one explain me a rigorous proof about when we can expect to have trivial mapping class group?



On trivial mapping class group of 3-manifolds
Dave Gabai [proved that the mapping class group of a closed hyperbolic 3-manifold is isomorphic to its isometry group](https://mathscinet.ams.org/mathscinet-getitem?mr=1354958). For a hyperbolic knot $K$ without any symmetries, for large enough $n$, $S^3\_{1/n}(K)$ will have a very short geodesic core by Thurston’s [hyperbolic Dehn filling theorem](https://en.wikipedia.org/wiki/Hyperbolic_Dehn_surgery). Hence any isometry of the unique hyperbolic metric (up to isotopy) on $S^3\_{1/n}(K)$ will preserve the core of the Dehn filling, and hence will preserve the knot complement and will be isotopic to the identity.


Addendum: A [new proof](https://arxiv.org/abs/1712.06197) that the mapping class group of a hyperbolic 3-manifold is isomorphic to the isometry group was proven by Bamler and Kleiner. They use Ricci flow-with-surgery to prove this (moreover, they prove that the space of hyperbolic metrics is contractible), whereas Gabai's proof uses a result that comes from an [intricate computer-assisted proof.](https://mathscinet.ams.org/mathscinet-getitem?mr=1973051)


