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Given any CAT(0) space $X$, we can define a map $s:X\times X\times [0;1]\rightarrow X$, such that $s(x,y,-)$ is the constant speed geodesic from $x$ to $y$ . Any isometry $f$ of $X$ is compatible with that map in the sense, that $s(f(x),f(y),t)=f(s(x,y,t))$. Then one can ask, whether any self-homeomorphism of $X$, which is compatible with $s$ in the upper sense is already a isometry.

This is clearly wrong for $X=\mathbb{R}^n$, as all affine maps are compatible with $s$. So the question is, whether these are the only examples.

For example I think I can show, that the $n$-dimensional hyperbolic space ($n\ge 2$) is rigid in that sense.

EDIT: Due to the big amout of counterexamples one could better ask the following question:

Are the spaces $\mathbb{R}^n$ the only spaces, which have self homeomorphisms compatible with $s$ (in the upper sense), that are not self-similarities ?

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  • $\begingroup$ Is "CAT(0)" a standard term? $\endgroup$
    – Gerald Edgar
    Commented Feb 19, 2010 at 14:43
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    $\begingroup$ "CAT(0)" is a standard term among differential geometers. It generalizes the notion of a Riemannian manifold with nonpositive curvature and originates from work of Alexandrov and Gromov. It's explained in wikipedia: en.wikipedia.org/wiki/CAT(k)_space $\endgroup$
    – Deane Yang
    Commented Feb 19, 2010 at 16:43

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The map which you call "geodesic preserving" is usually called "affine". It seems that affine maps to the real line are well understood even for general length space.

For your later edit: you may always take two spaces which admit self-similar maps and consider map on the product which move each coordinate with different coefficients.

A. Lytchak says that the following is a well known open question:

If such map exist then the space can be embedded into product of spaces and the map preserves product structure.

P.S. The "example" I gave before was not an example.

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If $Y$ is any CAT(0)-space, $X:=R^n \times Y$ will give another counterexample, so you might want to restrict to indecomposable $X$.
The next remark is that the union of the coordinate axes in $R^2$ gives another example, as does an arbitrary union of lines through the origin in $R^n$ with the induced length metric.

Another one in the same spirit is provided by the "forest" obtained by attaching a half-line to every point in $R^2$. Maybe it's true that that every self-similar CAT(0)-space $(X,d)$, i.e. one which is isometric to $(X,\rho \cdot d)$ for some $\rho \neq 1$, is a counter-example?

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  • $\begingroup$ Indeed, if you stitch together any collection of Euclidean spaces in a tree-like fashion, it will be a counterexample. $\endgroup$
    – Greg Kuperberg
    Commented Feb 19, 2010 at 14:40
  • $\begingroup$ We can take any one point union of Euclidean spaces to get a self-similar spaces. Note that then the only affine maps are these self-similarities. But I think there is a problem with the "forest". If I consider the geodesic, which goes first from (0,0) to (1,0) and then one up (in the direction of the half line). I guess the idea is, that every affine map of R² extends to a map of this space by preserving the distance in the attached directions. $\endgroup$
    – HenrikRüping
    Commented Feb 19, 2010 at 19:58
  • $\begingroup$ Applying (x,y)->(x+y,y) gives a geodesic, which is not parametrized by unit speed and hence the map is not compatible with s . Maybe I misunderstood how affine maps on R² should extend to this space. Hence also in this example there are only self-similarities. $\endgroup$
    – HenrikRüping
    Commented Feb 19, 2010 at 19:59

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