Let $\oplus$ denote the Minkowski sum in $\Bbb R^n$. For $k\in \mathbb N$, $k\ge 2$, define $$kA := \{ka: a\in A\}$$ for each $A\subset\Bbb R^n$. We know that if $A$ is convex, then $$kA = \underbrace{A \oplus \ldots \oplus A}_{k \text{ times}}$$ In general, we can only say that $$kA \supseteq \underbrace{A \oplus \ldots \oplus A}_{k \text{ times}}$$
Question: What conditions on $A$ are necessary and sufficient to ensure $$kA = \underbrace{A \oplus \ldots \oplus A}_{k \text{ times}}$$
It would also be interesting to know whether the set of conditions depend on $k$ or not.
Thanks!
If $A\subset \mathbb{R}^n$ is closed, then convexity is also necessary for the identity \begin{equation}\tag{1}\label{main-identity} kA=A\oplus\dots\oplus A. \end{equation}
To prove this, suppose we have a closed set $A$ which satisfies \eqref{main-identity}. We want to prove that $A$ is convex, so we fix two points $a,b\in A$ and want to prove that the points \begin{equation} c(t) := (1-t)a+tb,\quad 0\leq t\leq 1, \end{equation} are all contained in $A$.
For simplicity, we may consider first only the case $k=2$. In this case the identity \eqref{main-identity} implies that \begin{equation}\tag{2}\label{midpoint} x,y\in A\implies \frac{1}{2}x+\frac{1}{2}y\in A. \end{equation} That is the midpoint of two arbitrary points of $A$ is also contained in $A$.
Now consider the set \begin{equation} I_A = \{t\in[0,1]: c(t)\in A \} \end{equation} of values $t$ for which the corresponding point on the line segment between $a$ and $b$ is contained in the set $A$. By applying \eqref{midpoint} for points $c(t)$, we obtain \begin{equation}\tag{3}\label{param-midpoint} t,s\in I_A\implies \frac{1}{2}t+\frac{1}{2}s\in I_A. \end{equation}
By applying \eqref{param-midpoint} starting from $t=0$ and $s=1$ (corresponding to $x=a=c(0)$ and $y=b=c(1)$), and repeating with any successive points we obtain, we find that \begin{align} 0,1&\in I_A&&\implies& \frac{1}{2}&\in I_A\\ 0,\frac{1}{2}&\in I_A&&\implies& \frac{1}{4}&\in I_A\\ \frac{1}{2},1&\in I_A&&\implies& \frac{3}{4}&\in I_A\\ \frac{1}{2},\frac{3}{4}&\in I_A&&\implies& \frac{5}{8}&\in I_A \end{align} and so on, eventually finding that $I_A$ must contain all the dyadic rationals $i/2^m$, $i=0,\ldots,2^{m}$, $m\in\mathbb{N}$. Since we assumed that $A$ is closed, also $I_A\subset[0,1]$ is closed. Since the dyadic rationals are dense in $[0,1]$, we necessarily have that $I_A=[0,1]$, which proves the original convexity criterion for $A$ that we wanted in the $k=2$ case.
For the general $k\geq 2$ case, nothing important really changes. Using \eqref{main-identity} we obtain $x,y\in A\implies\frac{1}{k}(x+\dots+x+y)\in A$ and find an analogue of \eqref{param-midpoint} in the form \begin{equation} t,s\in I_A\implies \frac{k-1}{k}t+\frac{1}{k}s\in I_A. \end{equation} The only important thing here is that for any $t,s\in I_A$, we can construct another $u\in I_A$ with $t<u<s$. This guarantees that $I_A$ is dense in $[0,1]$ and we conclude as before.
For non-closed sets $A$ the only thing we can prove is that \begin{equation} \{\frac{i}{k^m}: i=0,\ldots,k^m, m\in\mathbb{N} \} \subset I_A \end{equation} but in general $I_A\neq[0,1]$ so we wouldn't have convexity. For instance we can do shenanigans involving rational equivalence classes under addition, such as taking $A\subset \mathbb{Q}^n$. But otherwise convexity is pretty much necessary and sufficient.