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 have proved 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.

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.

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 have proved 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 $$(*)\hspace{2mm} Z_{2^{\alpha_1}}\times Z_{2^{\alpha_2}}\times \cdots\times Z_{2^{\alpha_{k-1}}}\times Z_{2^{\alpha_k}}$$with $k\ge2$ and $1\leq\alpha_1\leq\alpha_2\leq\cdots\leq\alpha_{k-1}=\alpha_k$.

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.

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 have proved 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.