Let $(X,d)$ be a metric space and $x,y \in X$. Assume that for all $r > 0$ the balls $B_r(x)$ and $B_r(y)$ are isometric.
Is it true that there exists an isometry of $X$ sending $x$ to $y$?
No. Let $x$ and $y$ be connected by an edge and let's use the graph distance as our metric. At $x$, connect paths of length $n$ for each $n\in\mathbb N$. At $y$, do the same, but also connect an infinite path.
There is a 5-point example $\ X := \{x\ y\ a\ b\ c\},\ $ with (symmetric) metrics $\ d\ $ as follows:
$$d(x\ y) = d(a\ b) = 1$$
$$d(x\ a) = d(y\ b) = 2$$
$$d(x\ b) = d(y\ a) = 3$$
$$d(x\ c) = d(y\ c) = 6$$ $$d(a\ c) = 5\qquad\qquad d(b\ c) = 4$$
Oviously, $\ B_r(x)\ $ and $\ B_r(y)\ $ are isometric for every $\ r>0,\ $ while there is no isometry $\ f:X\rightarrow X\ $ for which $\ f(x)=y$.
REMARK 1 This example is ironic because while the respective balls are isometric, the isometry of the balls doesn't respect the centers. In this sense every bounded (and especially--finite) required example would be ironic.
REMARK 2 Number $\ 5\ $ is minimal.