Before I state the questions I have in mind, let me give some background. If one considers $S^2$ then it is known due to Kneser that $\textrm{Homeo}^{+}(S^2)$ has the homotopy type of $SO(3)$. By Smale's work we also know that $\textrm{Diff}^{+}(S^2)$ is homotopic to $SO(3)$. However, when we work in the homotopy category this changes. Later Hansen considered $\textrm{Aut}_0(S^2)$, the connected component of identity in the space of all self homotopy equivalences. He showed that its homotopy type is that of $SO(3)\times \mathbf{\Omega}$, where $\mathbf{\Omega}$ is the universal cover of the connected component of the constant loop in the double loop space $\Omega^2 S^2$.

Although this is a very nice and interesting fact, it's turns out to be hard to generalize his method of proof for higher spheres. For one, the homotopy type of $\textrm{Diff}(S^n)$ is not really known for $n\geq 7$ if I'm not mistaken. For another, Hansen crucially uses the fact that $S^2$ is the base of the usual fibration $SO(2)\to SO(3)\to S^2$ and the fact $SO(2)=S^1$ has no higher homotopy. This indubitably fails for higher spheres!

**Question 1** What is known about the homotopy type of $\textrm{Aut}_0(S^n)$ for $n\geq 3$?

I should say that the rational homotopy type is fairly easily calculable via Sullivan's minimal models. So, I'm looking for a bit more here.

**Question 2** What is the homotopy type of the identity component of self homotopy equivalences of $\vee_k S^2$, a bouquet of $2$-spheres?

Of course, one can ask this question for higher spheres but bearing in mind question 1, I decided I would be happy with an answer for $S^2$. If it helps, the homotopy groups of $\vee_k S^2$ can be calculated by Hilton-Milnor theorem. This fits in a long exact sequence of groups associated to $\textrm{Aut}_0^\ast(\vee_k S^2)\to \textrm{Aut}_0(\vee_k S^2)\to \vee_k S^2$ where the last map is evaluation of an automorphism at the common point of the bouquet and $\textrm{Aut}^\ast_0(\vee_k S^2)$ consists of based maps. But this doesn't seem to lead anywhere!

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