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Let $M^3$ denote the Poincare homology sphere. I am wondering what the possible orders of (smooth) automorphisms of $M$ are (I'm not sure if allowing arbitrary homeomorphisms changes things?). By presenting the space as $-1$-surgery on the trefoil, there is a automorphism of order 3 given by the 3-fold symmetry of the trefoil. Similarly, the framed link description of $M$ obtained by presenting $M$ as the 5-fold cyclic branched cover of $S^3$ branched over the trefoil has a 5-fold symmetry. Both of these symmetries can also be seen from the dodecahedron/icosahedron definition of $M$.

I believe that the $5k$-fold cyclic branched covers of $S^3$ branched over the trefoil are all $M$ for any $k \geq 1$ and therefore, by drawing the resulting framed link representations, we obtain automorphisms of order $5k$ for any $k$. I am not sure if any of these automorphisms are isotopic to the identity.

What are the possible orders of automorphisms of $M$? Which possible orders have an automorphism of that order that is not isotopic to the identity?

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    $\begingroup$ The answer to your question for all spherical 3-manifolds is given by the Rubinstein-McCullough paper. In that paper they compute the isometry groups of spherical 3-manifolds. You can use geometrization to argue any finite-order automorphism is conjugate to such an isometry. $\endgroup$ – Ryan Budney Sep 21 '18 at 4:00
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Let $G$ act smoothly and orientably. The quotient $M/G$ is a spherical 3-orbifold. Therefore, by the elliptization theorem, it may be given a metric of constant curvature 1; pulling back, then $M$ is given a constant curvature metric for which $G$ acts by isometries. (This was known to Thurston when the action of $G$ carried fixed points, and due to Perelman when $G$ acts freely.)

So we may just ask what the finite groups of isometries of $S^3/I$ is, where $I$ is the binary dihedral group. Consider the subgroup $\text{Isom}_I(S^3)$ of isometries of $S^3$ which normalize the (right) action of $I$, in the sense that $f(xi) = f(x)i'$ for some $i' \in I$. It is easy to see that every such function is a lift of an isometry of $S^3/I$, and there is a short exact sequence $1 \to I \to \text{Isom}_I(S^3) \to \text{Isom}(S^3/I) \to 1$.

Now recall that $\text{Isom}(S^3) = SO(4) = S^3 \times_{\pm 1} S^3$, the two factors acting by $(q_1, q_2) \cdot v = q_1 v q_2^{-1}$. The condition above on the action of $(q_1, q_2)$ becomes "For fixed $i$, there is an $i'$ so that $q_1 v i q_2^{-1} = q_1 v q_2^{-1} i'$ for all $v$." This is true if and only if $q_2 i q_2^{-1} = i'$ for every $i$. This implies $q_2 \in N_{SU(2)}(I) = I$.

Now we see that $\text{Isom}(S^3/I) = (S^3 \times_{\pm 1} I)/(1 \times I)$; the quotient is $S^3/\pm 1 = SO(3)$.

So every finite group of automorphisms is conjugate to the action of some subgroup of $SO(3)$, acting on $S^3/I = SO(3)/(I/\pm 1)$ by left multiplication. One concludes by listing all finite subgroups of $SO(3)$; this includes, of course, all cyclic groups.

The Smale conjecture (certainly known now for $\Sigma(2,3,5)$, but I don't know the history well enough to give the correct attribution) dictates that $\text{Isom}^+(M) \to \text{Diff}^+(M)$ is a homotopy equivalence when $M$ is a spherical 3-manifold. So $M$ has trivial mapping class group.

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  • $\begingroup$ I would be surprised if you couldn't extend this to the case of compact positive-dimensional group actions (it should be easier...), but I have put no effort into this. $\endgroup$ – Mike Miller Sep 20 '18 at 20:17

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