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3 corrected and simplified proof

Let $d_e$ be the usual metric on $\mathbb{R}^2$ and $S = \{(\frac1n,0) : n \operatorname{is} \operatorname{a} \operatorname{positive} \operatorname{integer}\}$. Let $(X,d)$ be $\mathbb{R}^2-S$ with the metric inherited from $(\mathbb{R}^2,d_e)$.

Clearly, $P_{(X,d)}(x,R,r)\leq P_{(\mathbb{R}^2,d_e)}(x,R,r)$ and $(\mathbb{R}^2,d_e)$ is packing-homogeneous. Consider a ball $B$ in $(\mathbb{R}^2,d_e)$ with center $C$, and disjoint balls of radius $r$ with centers $\{c_i : i\in I\}$ such that $\{c_i : i\in I\} \subseteq B_0$. Bounded subsets of $(\mathbb{R}^2,d_e)$ are totally bounded, so $|\{c_i : i\in I\}| < \infty$. Define $R(\theta)$ as the rotation of $(\mathbb{R},d_e)$ by $\theta$ radians about $C$, clearly these are all isometries which fixe $C$ and $B$ is invariant under. Since the points in $S$ are colinear, $|\{s\in S : d_e(C,c_i) = d_e(C,s)\}| \leq 2$, so if $C$ is not a member of $S$, then $|\{\theta\in (-\pi,\pi] : (r(\theta))(c_i)\in S\}| \leq 2$. This gives us $|\{\theta\in (-\pi,\pi] : (\exists i\in I)((r(\theta))(c_i)\in S)\}| =$ $|\displaystyle\bigcup_{i\in I} \{\theta\in (-\pi,\pi] : (r(\theta))(c_i)\in S\}| \leq |I|\cdot 2 < \infty\cdot 2 = \infty \leq |(-\pi,\pi]|$, so there exists $\theta\in (-\pi,\pi]$ such that $\neg (\exists i\in I)((r(\theta))(c_i)\in S)$. Let $\phi$ be such a $\theta$. Then $(\forall i\in I)((r(\phi))(c_i)\not\in S)$. This shows that if $C$ is not a member of $S$, then there is a packing in $B$ which does not use any points of $S$, so $P_{(\mathbb{R}^2,d_e)}\leq P_{(X,d)}(x,R,r)$. Let $x,y$ be members of $X$, then $P_{(X,d)}(x,R,r)\leq P_{(\mathbb{R}^2,d_e)}(x,R,r) = P_{(\mathbb{R}^2,d_e)}(y,R,r) \leq P_{(X,d)}(y,R,r)$ $\leq P_{(\mathbb{R}^2,d_e)}(y,R,r) = P_{(\mathbb{R}^2,d_e)}(x,R,r)\leq P_{(X,d)}(x,R,r)$, which shows that the packing function on $(X,d)$ is independent of basepoint. Therefore $(X,d)$ is packing-homogeneous.

$(X,d)$ is not locally simply connected compact near $(0,0)$, but it is locally simply connected compact near $(0,1)$, so $(X,d)$ is not even topologically homogeneous. $(\mathbb{R}^2,d_e)$ is connected and To get to $S$ is nowhere dense (0,0)$from any point in it,$(X,d)$, if you start below the x axis travel down to$y=-1$, otherwise travel up to$y=1$, then travel horizontally to$x=0$, then travel vertically to$y=0$. This shows that every point has a path to$(0,0)$, so$(X,d)$is also connected. Therefore$(X,d)$is a connected packing-homogeneous space which is not homogeneous. QED Demanding completeness might be enough. It would certainly stop my idea. 2 more general formatting Let$d_e$be the usual metric on$\mathbb{R}^2$and $S = {(\frac1n,0) \{(\frac1n,0) : n \operatorname{is} \operatorname{a} \operatorname{positive} \operatorname{integer}}$. operatorname{integer}\}$. Let $(X,d)$ be $\mathbb{R}^2-S$ with the metric inherited from $(\mathbb{R}^2,d_e)$.

Clearly, $P_{(X,d)}(x,R,r)\leq P_{(\mathbb{R}^2,d_e)}(x,R,r)$ and $(\mathbb{R}^2,d_e)$ is packing-homogeneous. Consider a ball $B$ in $(\mathbb{R}^2,d_e)$ with center $C$, and disjoint balls of radius $r$ with centers $\{c_i : i\in I\}$ such that $\{c_i : i\in I\} \subseteq B_0$. Bounded subsets of $(\mathbb{R}^2,d_e)$ are totally bounded, so $|\{c_i : i\in I\}| < \infty$. Define $R(\theta)$ as the rotation of $(\mathbb{R},d_e)$ by $\theta$ radians about $C$, clearly these are all isometries which fixe $C$ and $B$ is invariant under. Since the points in $S$ are colinear, $|\{s\in S : d_e(C,c_i) = d_e(C,s)\}| \leq 2$, so if $C$ is not a member of $S$, then $|\{\theta\in (-\pi,\pi] : (r(\theta))(c_i)\in S\}| \leq 2$. This gives us $|\{\theta\in (-\pi,\pi] : (\exists i\in I)((r(\theta))(c_i)\in S)\}| = |\displaystyle\bigcup_{i\in$ $|\displaystyle\bigcup_{i\in I} \{\theta\in (-\pi,\pi] : (r(\theta))(c_i)\in S\}| \leq |I|\cdot 2 < \infty\cdot 2 = \infty \leq |(-\pi,\pi]|$, so there exists $\theta\in (-\pi,\pi]$ such that $\neg (\exists i\in I)((r(\theta))(c_i)\in S)$. Let $\phi$ be such a $\theta$. Then $(\forall i\in I)((r(\phi))(c_i)\not\in S)S)$. This shows that if $C$ is not a member of $S$, then there is a packing in $B$ which does not use any points of $S$, so $P_{(\mathbb{R}^2,d_e)}\leq P_{(X,d)}(x,R,r)$. Let $x,y$ be members of $X$, then $P_{(X,d)}(x,R,r)\leq P_{(\mathbb{R}^2,d_e)}(x,R,r) = P_{(\mathbb{R}^2,d_e)}(y,R,r) \leq P_{(X,d)}(y,R,r)\leq P_{(X,d)}(y,R,r)$ $\leq P_{(\mathbb{R}^2,d_e)}(y,R,r) = P_{(\mathbb{R}^2,d_e)}(x,R,r)\leq$P_{(X,d)}(x,R,r), P_{(X,d)}(x,R,r)$, which shows that the packing function on$(X,d)$is independent of basepoint. Therefore$(X,d)$is packing-homogeneous.$(X,d)$is not locally simply connected near$(0,0)$, but it is locally simply connected near$(0,1)$, so$(X,d)$is not even topologically homogeneous. (\mathbb{R}^2,d_e)$(\mathbb{R}^2,d_e)$is connected and$S$is nowhere dense in it, so$(X,d)$is also connected. Therefore$(X,d)$is a connected packing-homogeneous space which is not homogeneous. QED Demanding completeness might be enough. It would certainly stop my idea. 1 Let$d_e$be the usual metric on$\mathbb{R}^2$and$S = {(\frac1n,0) : n \operatorname{is} \operatorname{a} \operatorname{positive} \operatorname{integer}}$. Let$(X,d)$be$\mathbb{R}^2-S$with the metric inherited from$(\mathbb{R}^2,d_e)$. Clearly,$P_{(X,d)}(x,R,r)\leq P_{(\mathbb{R}^2,d_e)}(x,R,r)$and$(\mathbb{R}^2,d_e)$is packing-homogeneous. Consider a ball$B$in$(\mathbb{R}^2,d_e)$with center$C$, and disjoint balls of radius$r$with centers $\{c_i : i\in I\}$ such that $\{c_i : i\in I\} \subseteq B_0$. Bounded subsets of$(\mathbb{R}^2,d_e)$are totally bounded, so $|\{c_i : i\in I\}| < \infty$. Define$R(\theta)$as the rotation of$(\mathbb{R},d_e)$by$\theta$radians about$C$, clearly these are all isometries which fixe$C$and$B$is invariant under. Since the points in$S$are colinear, $|\{s\in S : d_e(C,c_i) = d_e(C,s)\}| \leq 2$, so if$C$is not a member of$S$, then $|\{\theta\in (-\pi,\pi] : (r(\theta))(c_i)\in S\}| \leq 2$. This gives us $|\{\theta\in (-\pi,\pi] : (\exists i\in I)((r(\theta))(c_i)\in S)\}| = |\displaystyle\bigcup_{i\in I} \{\theta\in (-\pi,\pi] : (r(\theta))(c_i)\in S\}| \leq |I|\cdot 2 < \infty\cdot 2 = \infty \leq |(-\pi,\pi]|$, so there exists$\theta\in (-\pi,\pi]$such that$\neg (\exists i\in I)((r(\theta))(c_i)\in S)$. Let$\phi$be such a$\theta$. Then$(\forall i\in I)((r(\phi))(c_i)\not\in S). This shows that if $C$ is not a member of $S$, then there is a packing in $B$ which does not use any points of $S$, so $P_{(\mathbb{R}^2,d_e)}\leq P_{(X,d)}(x,R,r)$. Let $x,y$ be members of $X$, then $P_{(X,d)}(x,R,r)\leq P_{(\mathbb{R}^2,d_e)}(x,R,r) = P_{(\mathbb{R}^2,d_e)}(y,R,r) \leq P_{(X,d)}(y,R,r)\leq P_{(\mathbb{R}^2,d_e)}(y,R,r) = P_{(\mathbb{R}^2,d_e)}(x,R,r)\leq$P_{(X,d)}(x,R,r), which shows that the packing function on $(X,d)$ is independent of basepoint. Therefore $(X,d)$ is packing-homogeneous.

$(X,d)$ is not locally simply connected near $(0,0)$, but it is locally simply connected near $(0,1)$, so $(X,d)$ is not even topologically homogeneous. (\mathbb{R}^2,d_e) is connected and $S$ is nowhere dense in it, so $(X,d)$ is also connected. Therefore $(X,d)$ is a connected packing-homogeneous space which is not homogeneous. QED

Demanding completeness might be enough. It would certainly stop my idea.