It is well known that the automorphisms of a group $G$ form a group under composition, and that the group of inner automorphisms $\phi (x)=gxg^{1}$ forms a normal subgroup of $\mbox{Aut}(G)$. Thus, $\mbox{Aut}(G)$ is simple if and only if either $\mbox{Inn}(G)=\mbox{Aut}(G)$ or $\mbox{Inn}(G)$ is trivial. In the second case, since $G/Z(G)=\mbox{Inn}(G)$, $G$ must be abelian. My question is, when does $\mbox{Inn}(G)=\mbox{Aut}(G)$? Or, as it is unlikely that the general case is not fully understood, are there nice classes of groups for which there are a nice set of criteria for $\mbox{Inn}(G)=\mbox{Aut}(G)$.

Here is an approximation of an answer to "For what finite groups is Aut(G) simple?" As Daniel Miller mentioned, Inn(G) is a normal subgroup of Aut(G), so for Aut(G) to be simple either Inn(G) = 1, in which case G is abelian, or Inn(G) = Aut(G) is simple. The former case should be somewhat easy to handle assuming G is finite. In the latter case, we have that G/Z(G) is simple. If G is also perfect, then G is called quasisimple. Of course, G need not be perfect as G ≅ A_{5} × 2 shows. However, I believe this is the only obstruction, so ignoring a possible cyclic direct factor of order 2, G/Z(G) is simple, and G is quasisimple. The finite quasisimple groups and their automorphism groups are classified, but the classification is a bit long. For a fixed simple group, X = G/Z(G), there are only finitely many isomorphism classes of quasisimple groups D such that D/Z(D) = X. In fact there is a unique largest one called the Schur cover, that I'll call D. If Z(D) is cyclic, then in fact Aut(G) = Aut(X) = Aut(D) does not pay any attention to the center. So all we need to do is find all X with Aut(X) = X [and each one works], and all X with Z(D) noncyclic [and check which ones work]. Having done most, but not all, of that, I thought it might help to record the basic result: If G = H×T where T=1 if H is abelian and T is cyclic of order dividing 2 otherwise, and where H is on the following list, then Aut(G) is simple:
Additionally if Aut(G) is simple, then G = H×T as above, except possibly H/Z(H) is on the following list:
These are groups with noncyclic multiplier other than Sz(8) [definitely an example] and Ω^{+}(8,2) [not an example]. The Ω^{+}(4n,q) case should be mostly easy, as there are too many automorphisms to kill. The others would be easy in an ideal world, but as far as I know our computational knowledge of these groups is limited and/or flawed. Of course, I also need to check the abelian case carefully, but I think 3,4,6 and 2^n are the only abelian examples. It would make another good answer: For what torsion abelian groups G is Aut(G) simple? This would handle the abelian groups here, as well as some of the original poster's interest, without delving into the nastier aspects of abelian groups. 


Obraztsov has shown that if $p$ is a sufficiently large prime, then there exists a finitely generated infinite simple complete group $G$, all of whose proper subgroups are cyclic of order $p$. In particular, $G$ is an example of a group such that $Aut(G)$ is an infinite simple group. The relevant reference is: V. N. OBRAZTSOV, `On infinite complete groups', Comm. Algebra 22 (1994) 58755887 


This is not an answer to your exact question (which I interpreted to be 'When does $\mathrm{Inn}(G)=\mathrm{Aut}(G)$?'as pointed out in the comments, this is not the same as asking for $\mathrm{Aut}G$ to be simple), and is only really interesting if you care about examples where $G$ is infinite. If you do care about $G$ infinite, then a natural slight weakening is to ask for criteria for $\mathrm{Out}(G)$ to be finite. One such criterion is provided by Paulin's Theorem. Theorem. If $G$ is wordhyperbolic and $\mathrm{Out}(G)$ is infinite then $G$ splits (as an amalgamated free product or HNN extension) over a virtually cyclic subgroup. It is known that, using some suitable definition of 'randomly chosen', a randomly chosen finitely presented group is torsionfree, wordhyperbolic and does not split. So one can conclude that a 'randomly chosen' finitely presented group $G$ is of finite index in its automorphism group. 


Does anyone know of any infinite simple groups which are the Automorphism groups of other groups? E.g. some $G$ with $Aut(G)$ infinite and simple? 


let $G$ be non ableian group and $A$ be set of all groups including $Z(G)$. for All H in A,send H to Z(H)(Notice that this is a map from A to A). notice if Z(H)=Z(G) all H in A, it cause a contradiction(easy to show) if Inn(G)=Aut(G) then there is a uniqe proper group with Z(H)=Z(G) in A. M.Y.K 

