Homotopy type of the self-homotopy equivalences of a bouquet of spheres 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!
 A: Regarding question 1 the following is  well known.
For a reasonably nice space  $X$ (say, a finite connected CW-complex) let $Aut_0(X)$ be the monoid of self-equivalences of $X$ homotopic to the identity and let $Aut_0^\bullet(X)$ be the submonoid fixing a basepoint. Then there is a natural fibration given by the evaluation map $Aut_0^\bullet(X)\to Aut_0(X)\to X$. This gives rise to the fibration $X\to BAut_0^\bullet(X)\to BAut_0(X)$ which is the universal fibration with fiber $X$.
For $X=S^n$ it's easy to see that $Aut_0^\bullet(S^n)\cong Map_0^\bullet(S^n, S^n)\cong Map_0^\bullet(S^{n-1}, \Omega S^n)\cong \ldots Map_0^\bullet (S^0, \Omega^n(S^n))=\Omega^n(S^n)$. In particular, $\pi_k(Aut_0^\bullet(S^n))\cong \pi_{n+k}(S^n)$. Combining this with the fibration $Aut_0^\bullet(S^n)\to Aut_0(S^n)\to S^n$ this in principle gives you homotopy groups of $Aut_0(S^n)$. In practice though I don't think this is very useful away from rational  coefficients or for small values of $k$ where you have stability.
I'm not familiar with the general theory for spaces other than spheres. As I mentioned in my comment, rationally the situation is in principle  well understood. But even for rational coefficients the homotopy structure of $Aut_0(X)$ is complicated when $X$ is a wedge of spheres.  As I mentioned, rational homotopy groups of $Aut_0(X)$ grow exponentially. See "The monoid of self-homotopy equivalences of some homogeneous spaces" by  Félix and Thomas for a careful proof of that.
One general fact that I know is a result of Dror, Dwyer and  Kan  that when $X$ is a nilpotent finite $CW$-complex then   $BAut_0(X)$ has finite type, i.e. it has the homotopy type of a CW complex with a finite number of cells in each dimension.
