Which are the finite groups $(G,\cdot)$ with the following property: for every $f \in Aut(G)$, there are $g,h\in Aut(G)$ such that $f(x)=g(x)\cdot h(x), \forall x\in G?$ 


I already have verified that:
 (i) such a group is abelian;
 (ii) all finite abelian groups of odd order have this property;
 (iii) a finite abelian 2-group satisfying this property must be of type $$(\*)\hspace{2mm} \mathbb{Z}\_{2^{\alpha\_1}}\times \mathbb{Z}\_{2^{\alpha\_2}}\times \cdots\times \times \mathbb{Z}\_{2^{\alpha\_k}}$$where each $\alpha\_i$ occurs at least twice.


In my opinion, the above class consists of all direct products $G\_1\times G\_2$, where $G\_1$ is an abelian 2-group of type $(\*)$ and $G\_2$ is an abelian group of odd order, but I failed to prove it. 



A question on automorphisms of finite abelian groups
Here is an outline for odd order abelian $p$-groups:


The main point is that every non-zero element in a field with at least three elements is a sum of two non-zero elements. Using this, you can show that irreducible matrix in $GL\_n(\mathbf Z/p\mathbf Z)$ can be written as a sum of two non-singular matrices, for such an irreducible matrix corresponds to multiplication by a primitive element of the field of order $p^n$.


Now, using the Jordan canonical form matrices over a finite field (see Theorem A.22 in [my notes](http://www.imsc.res.in/~amri/html_notes/notesap1.html#x8-32000A.4)) you can show that every non-singular matrix over $\mathbf Z/p\mathbf Z$ is a sum of two non-singular matrices.


Finally, the endomorphism algebra of a finite abelian $p$-group modulo is radical is a product of matrix groups over $\mathbf Z/p\mathbf Z$ (see section 6 of [this paper](http://arxiv.org/abs/math/0611897)). So any invertible element of the endomorphism algebra can be written as a sum of two invertible elements modulo the radical of this ring. But adding something in the radical does not affect invertibility.


