What polygons can be shrunk into themselves? Let's call a polygon $P$ shrinkable if any down-scaled (dilated) version of $P$ can be translated into $P$. For example, the following triangle is shrinkable (the original polygon is green, the dilated polygon is blue):

But the following U-shape is not shrinkable (the blue polygon cannot be translated into the green one):

Formally, a compact $\ P\subseteq \mathbb R^n\ $ is called shrinkable iff:
$$\forall_{\mu\in [0;1)}\ \exists_{q\in \mathbb R^n}\quad \mu\!\cdot\! P\, +\, q\ \subseteq\ P$$
What is the largest group of shrinkable polygons?
Currently I have the following sufficient condition: if $P$ is star-shaped then it is shrinkable. 
Proof: By definition of a star-shaped polygon, there exists a point $A\in P$ such that for every $B\in P$, the segment $AB$ is entirely contained in $P$. Now, for all $\mu\in [0;1)$, let $\ q := (1-\mu)\cdot A$. This effectively translates the dilated $P'$ such that $A'$ coincides with $A$. Now every point $B'\in P'$ is on a segment between $A$ and $B$, and hence contained in $P$.

My questions are:
A. Is the condition of being star-shaped also necessary for shrinkability?
B. Alternatively, what other condition on $P$ is necessary?
 A: Let $\ L\ $ be a Hilbert space. Let $\ P\subseteq L\ $ be a non-empty compact subset. Then $\ P\ $ is called $\ \mu$-shrinkable $\ \Leftarrow:\Rightarrow$
$$\exists_{q\in L}\ \ \mu\cdot P\ +\ q\ \subseteq\  P$$
for arbitrary $\ \mu\ge 0\ $ (thus $\ \mu \le 1\ $ when $\ |P|>1$).
Let $\ m(P)\ $ be the set of all $\ \mu\ge 0\ $ such that $\ P\ $ is $\ \mu$-shrinkable. Following @E.S.Halevi, let $\ P\ $ be called shrinkable $\ \Leftarrow:\Rightarrow\ \ m(P) = [0;1].\ $ Then

THEOREM  The following three properties of a non-empty compact $\ P\subseteq L\ $are equivalent:
  
  
*
  
*P is a star set;
  
*P is shrinkable;
  
*$\ \sup (\ m(P)\cap[0;1)\ )\ =\ 1$
  

PROOF   Implications $\ 1\Rightarrow 2\Rightarrow 3\ $ are trivial. we need only $\ 3\Rightarrow 1.\ $ Thus assume condition $3$.
Consider map $\ f_\mu : x\mapsto \mu\cdot x + q_\mu,\ $ of $\ P\ $ into itself, for every $\ \mu\in m(P)\cap[0;1).\ $ Then by Banach's fpp there exists a unique $c_\mu\in P\ $ such that $\ c_\mu = \mu\cdot c_\mu + q_\mu,\ $ so that $\ q_\mu = (1-\mu)\cdot c_\mu.\ $ Thus there is a limit point $\ c_1\in P\ $ of a certain sequence of points $\ c_\mu\ $ for which $ \lim \mu = 1$.
Observe that for $\ \nu:=\mu^k\ $ the composition $\ g_\nu:=\bigcirc^k f_\mu\ $ has the same fixed point (I am going to choose at the most one $\ k\ $ for each $\ \mu;\ $ also
$$\forall_{x\in L}\ \ g_\nu(x)\ =\ \nu\cdot x\ +\ (1-\nu)\cdot c_\mu$$
Now let's consider an arbirary $\ \kappa\in[0;1).\ $ I'll show that function
$$\ F_\kappa\ :\ x\ \mapsto\ \kappa\cdot (x-c_1)+c_1\ \ =\ \ \kappa\cdot x\ +\ (1-\kappa)\cdot c_1$$
maps $\ P\ $ into itself (for every such $\kappa,\ $ so that will be the end of the proof).
Thus let $\ \epsilon > 0.\ $ Then there exist $\ \mu\in[0;1)\ $ and natural $\ k,\ $ such that $\ |c_\mu-c_1|<\epsilon\ $ and $\ |\nu-\kappa|<\epsilon\ $ for $\ \nu:=\mu^k,\ $ hence for arbitrary $\ x\in P$:
$$ |g_\nu(x)-F_\kappa(x)|\ \le\ |g_\nu(x)-F_\nu(x)|\ +\ |F_\nu(x)-F_\kappa(x)|$$
where
$$\ |F_\nu(x)-F_\kappa(x)|\ =\ |(\nu-\kappa)\cdot x + (\kappa-\nu)\cdot c_1|\ =\ |\nu-\kappa|\cdot|x-c_1|$$ 
henceforth
$$|F_\nu(x)-F_\kappa(x)|\ \le\ \epsilon\cdot |x-c_1|$$
Next
$$|g_\nu(x)-F_\nu(x)|\ =\ |(1-\nu)\cdot c_\mu - (1-\nu)\cdot c_1|\ =\ |1-\nu|\cdot|c_\mu-c_1|\ \le\ \epsilon$$
These inequalities imply:
$$ |g_\nu(x)-F_\kappa(x)|\ \le\ (|x-c_1|+1)\cdot\epsilon$$
or
$$ |g_\nu(x)-F_\kappa(x)|\ \le\ D\cdot\epsilon$$
where  $\ D := diam(P)$
Thus for every $\ \delta > 0\ $ let $\ \epsilon:=\frac\delta D\ $ such that... OK, enough of this $\delta$-$\epsilon$ business, $\ F_\kappa(x)\in P$.
END of PROOF

REMARK The theorem holds not just for the Hilbert spaces but also for Banach spaces. One should be also able to replace translations by arbitrary linear isometries. I am even curious and hopeful about considering this kind of theorems for the locally convex linear spaces.

A: Any simply connected polygon must be star-shaped to be shrinkable. I have made minor edits below to treat the more general case.
Let $D$ be a polygon with convex hull $H$. Assume we are given a non-trivial shrinking of $D$; view this as a map from $H$ to itself. This map must have a fixed point $x$, either by algebraic topology or an iterative construction. 
This means it suffices to consider only dilations centered at a point $x$ in $H$, rather than dilations followed by translations.
For any $x$, if there is a point $y$ in $D$ so that the segment from $x$ to $y$ is not contained in $D$, then a $(1-\epsilon)$-dilation of $H$ centered at $x$ will not carry $D$ into $D$ for any positive $\epsilon$ smaller than some $\epsilon(x)>0$. If $D$ is not star-shaped, take the minimum $\delta$ of $\epsilon(x)$ over $x\in H$, and then no $(1-\delta)$-dilation of $H$ centered at a point in $H$ carries $D$ into $D$.
A: Here is my variant, a bit more geometrical.
Denote by $P$ the original polygon, and $P_\lambda$ the contracted polygon with a factor $\lambda \in (0,1)$ which lies inside $P$. Note that $P_\lambda$ is obtained from $P$ after a dilation and a translation, and therefore there exists a point $O_\lambda$ such that $P_\lambda$ is the image of $P$ under a homothety $h_\lambda$ of center $O_\lambda$ and factor $\lambda$.
Now we know that $h_\lambda : P \to P_\lambda \subset P$ is well defined, continuous and a contraction. Therefore, since $P$ is closed, $h_\lambda$ has a fixed point in $P$, which can only be $O_\lambda$. As a consequence $O_\lambda \in P\cap P_\lambda$.
Pick a sequence $\lambda_n \to 1$ and denote $O_n = O_{\lambda_n}$. Since $(O_n)$ is contained in a compact set $P$, it follows that it has at least one limit point $O$. For keeping the notations simple, we assume the whole $(O_n)$ is convergent to $O$. 
Take now $X \in P$ and assume that $[OX]$ is not contained in $P$. Then there exists $Y \in [OX] \setminus P$. Since $P$ is closed, there exists a maximal open subsegment $(X_1X_2)$ of $[OX]$ which contains $Y$ and is out of $P$ ($X_1 \in (OY)$). Obviously $X_1,X_2 \in P$. Moreover, there exists an open cone $C$ of direction given by $(X_1X_2)$, with angle and length $\varepsilon$ small enough, which contains $(X_1X_2)\cap B(X_2,\varepsilon)$ and does not intersect $P$. This happens since $P$ is a polygon and the exterior of $P$ near $X_2$ is either an angle or a half-plane. Consider now $Z_n = h_{\lambda_n}(X_2)$. By hypothesis we have $(Z_n) \subset P$. 
Since $O_n \to O$, for $n$ large enough the angle $\angle O_nX_2O$ will be smaller than $\varepsilon/2$. Since $O_nZ_n = \lambda_n O_nX_2$ and $\lambda_n \to 1$ the points $Z_n$ will lie in the cone $C$ for $n$ large enough. This contradicts the fact that $Z_n$ is in $P$.
Therefore $P$ is star-shaped with respect to $O$.
A: If I understand the question correctly the requirement is for a figure, F, such there exists a translation T(c) for all contractions c, such that cF+T(c) lies within F. It seems to me that that criterion holds for monoconvex hexagons (chevrons) and biconvex hexagons (hourglasses), which are not stars.
