Do all metric spaces satisfy this property (transitive action by isometries). Do all metric spaces satisfy this property? Suppose $A$ and $B$  are finite sequences $a_1,a_2\dots a_n$ and $b_1,b_2\dots b_n$ such that $d(a_i,a_k)=d(b_i,b_k)$ for all valid $i,k$. We say a metric space is transitive if whenever two sets $A$ and $B$ satisfy the  previous property there is an isometry sending $a_i$ to $b_i$ for all valid $i$. Clearly $\mathbb R$ satisfies this property, and I am fairly certain $\mathbb R^2$ does too, do all metric spaces satisfy this, does $\mathbb R^2$?
Thank you very much in advance.
Regards.
 A: 1) No (to the question in the subject line and the first sentence of the body). E.g., the subspace $\{0, 1, 2\}$ of $\mathbb{R}$ has no isometry that maps $0$ to $1$ and $1$ to $0$.
2) Yes (to the other question tagged on at the end) $\mathbb{R}^2$ is what you call transitive. To see this note that it is enough to consider the case when $n = 3$ (because the distance of a point from the vertices of a triangle fixes that point).
A: This is an amazing property of Euclidean geometry (and others, like hyperbolic and spherical geometry).  Given an isometry mapping some subset $A$ onto subset $B$, it may be extended to an isometry mapping the whole space onto itself.  But most metric spaces fail to have even the simplest version of this, as Rob showed in his example.
A: This is true in $\mathbb R^n$ (though not in other spaces, as others have pointed out). In particular, let us work in an inner product space and prove the statement that, if for a basis $A$ and a sequence $B$, if $\|a_i\|=\|b_i\|$ and $\|a_i-a_j\|=\|b_i-b_j\|$ then the unique linear map $f$ satisfying $f(a_i)=b_i$ is an isometry. Notice that this implies that $b_i\cdot b_j=a_i\cdot a_j$, since we can write
$$b_i\cdot b_j=\frac{\|b_i\|^2+\|b_j\|^2-\|b_i-b_j\|^2}2=\frac{\|a_i\|^2+\|a_j\|^2-\|a_i-a_j\|^2}2=a_i\cdot a_j.$$
The proof from here is easy - we just need to show that $v\cdot v = f(v)\cdot f(v)$ for all $v$. To do this, use that $A$ is a basis to write $$v=\sum_{i\in I}c_i\cdot a_i$$
$$f(v)=\sum_{i\in I}c_i\cdot b_i$$
for some real coefficients $c_i$. Then, we can see that $$v\cdot v=\sum_{i\in I}\sum_{j\in I}c_ic_ja_i\cdot a_j$$
$$f(v)\cdot f(v)=\sum_{i\in I}\sum_{j\in I}c_ic_jb_i\cdot b_j$$
However, since $a_i\cdot a_j=b_i\cdot b_j$, the expressions are equivalent and this is an isometry.
To extend this to the original problem, simply notice that if we take some subset $I$ of the indexes of the sequence $A$ such that every element in $A$ is an affine combination of the $a_i$ for $i\in I$, but that no $a_i$ is an affine combination of the other $a_i$ (i.e. it is a minimal affine span), then, by translating an element to the origin, we can immediately prove that there is an isometry taking the affine span of $A$ to the affine span of $B$, mapping $a_i$ to $b_i$. Then, we can easily, by mapping some orthogonal basis including the vectors in the span of $A$ to a similar basis including the vectors in the span of $B$, extend that isometry to the whole space.
