TITLE: If all balls around two points are isometric... -- manifold version
QUESTION [16 upvotes]: This question is a natural follow-up of this other question, asked earlier today by wspin.
Let's say that a metric space $(X,d)$ has two poles if:

there are two distinct points $x$, $y$ such that for every $r\ge 0$ the open balls $B_r(x)$ and $B_r(y)$ are isometric;
there is no isometry $f$ of $(X,d)$ such that $f(x)=y$.

Bjørn Kjos-Hanssen gave a very pretty example of a space with two poles (it's the metric space associated to an infinite graph). I've tried to turn this example into a surface by "inflating" it (imagine it in the 3-space, and take the boundary of a regular neighourhood), but failed. Hence my question:

Are there connected Riemannian manifolds with two poles?

If one drops the connectedness assumption, then Włodzimierz Holsztyński's answer to the question linked above is a 0-dimensional example. One can also ask for more: compactness, trivial $\pi_1$, curvature restrictions, etc., but I'll stick with the broader question for now.
The rough meta-question in the back of my mind is: "How nice can a bipolar space be?", so answers/comments in this direction are very welcome.

REPLY [12 votes]: The answer is no if $M$ is assumed to be complete: 
Let $x, y \in M$ such that for all $r > 0$ the balls $B_r(x)$ and $B_r(y)$ are isometric via 
$$f_r : B_r(x) \to B_r(y).$$ 
Set $R := \inf \{ r \mid B_r(x) = M\}$ (here $R = \infty$ if $M$ is unbounded). 
First I claim that 
For all $0 < r < R$ we have $f_r(x) = y$:
Assume on the contrary that $f_r(x) \neq y$. Let $z \in M \setminus \overline B_r(y)$ and $\gamma_1 : [0,l] \to M$ be a minimal arc length geodesic from $y$ to $z$. Then $\gamma_1(r) \in \partial B_r(y)$. Also let $\gamma_2$ be a minimal geodesic from $f_r(x)$ to $\gamma_1(r)$. Since $f_r^{-1}(\gamma_1(r)) \in \partial B_r(x)$ (here we extend $f$ to $\overline B_r(x)$ by metric completion), $f_r^{-1}(f_r(x)) = x$, $\gamma_2$ is minimal and $f_r$ is an isometry it follows that $\gamma_2$ has length $r$. But then also the curve obtained by concatenating $\gamma_2$ and ${\gamma_1}_{\vert [r,l]}$ is minimal between $f_r(x)$ and $z$. This is a contradiction, since geodesics do not branch.
Now, consider a sequence $f_{r_n}$ of such isometries with $f_{r_n}(x) = y$ and ${r_n} \to R$ for $n \to \infty$. After passing to a subsequence we may assume that $f_{r_n}$ converges to an isometry $f : M \to M$ with $f(x) = y$ (the isometries are determined by their differentials at $x$, which converge after some subsequence by Ascoli-Arzelà (or simply by linearity)).
Two remarks:

This argument holds more generally for complete length spaces with curvature bounded below. 
I am not sure whether the completeness assumption is necessary.