Are these exactly the abelian groups? I'm thinking about the following condition on a group $G$.

$$(\forall A\subseteq G)(\forall g\in G)(\exists h\in G)\ Ag=hA.$$

Obviously every abelian group $G$ satisfies this condition. Are there any other groups that do? Can we give a familiar characterization for them? Can we give one if we confine the considerations to finite groups?
Certainly not all groups satisfy the condition. Let $G$ be the free group on $\{x,y,z\}.$ Let $A=\{x,y\}$ and $g=z.$ Then $$Ag=\{x,y\}z=\{xz,yz\}.$$ Suppose there is $h\in G$ such that $\{hx,hy\}=hA=\{xz,yz\}.$ Then either $$\begin{cases}hx=xz\\hy=yz\end{cases}$$
or $$\begin{cases}hx=yz\\hy=xz\end{cases}$$
From the first case we get $h=xzx^{-1}$ and $h=yzy^{-1}$, which is a contradiction. From the second case we get $h=yzx^{-1}$ and $h=xzy^{-1}$, which is also a contradiction.
 A: Assume $ab\ne ba$. Let $A=\{1,a\}$, $g=b$. Then there is $h\in G$ such that $\{h,ha\}=\{b,ab\}$. This needs $h=b\lor h=ab$.
In the first case $ha=ba\ne ab$, so this fails.
Therefore $h=ab$ and $ha=aba=b$. Similarly, $bab=a$. This implies $aa=abab=bb$. We conclude
$$a=bab=bbabb=aaaaa, $$
hence $a^4=1$ and similarly $b^4=1$.
Now take $A=\{1,a,b\}$ and $g=b$. Then there is $h\in G$ such that $\{h,ha,hb\}=\{b,ab,b^2\}$.


*

*$h=b$: Then $hb=b^2$ implies $ba=ha=ab$, contradiction

*$h=ab$: Then $ha=aba=b$ implies $hb=b^2$, i.e. $a=1$ and of course $ab=ba$, contradiciton.

*$h=b^2=a^2$: Then $ha=a^3=a^{-1}\ne b$, hence $ha=ab$, i.e. $a^2=b$ and of course $ab=ba$, contradiction

A: Suppose $G$ satisfies your condition. Let $x,y \in G$ be distinct elements of $G$ different from $1$ so that the listed elements of all sets in this answer are distinct. Pick $g=x$ and $A=\{1,x,y\}$ Then $Ag = hA$ implies
$$ \{ x, xx, yx \} = \{ h, hx, hy \} $$
Case 1: $x=h$
Then $\{ xx,yx \} = \{ xx,xy \}$, and $yx = xy$
Case 2: $x = hx$
Then $h=1$ and $\{ xx, yx \} = \{ 1, y \}$.
If $yx = y$, then $x=1$ and $xy=yx$.
Otherwise $yx=1$, and so $y=x'$ and $xy=1=yx$.
Case 3: $x=hy$
Then $h=xy'$ and $\{ xx, yx \} = \{xy',xy'x\}$
If $xx = xy'$, then $x=y'$ and thus $xy=1=yx$.
Otherwise $xx = xy'x$ and thus $1=y'$, and again $xy=yx$.
In all cases where $x,y,1$ are distinct, we've shown $x$ and $y$ commute. Thus $G$ is an abelian group.
