Let $X$ be an n point finite set equipped with a metric $d$. An isometry is a map $\varphi$ $:X \mapsto X$ satisfying that $d(\varphi(x),\varphi(y))=d(x,y)$ for any $x,y \in X$. Identify $X$ with the set {$1,2,...,n $}. We can say every isometry is an element of $S_n$(the permutation group of n elements) and the isometry group of $X$ (denote it by $Iso(X,d)$) is a subgroup of $S_n$. The question is the following: if there is no isometry moving $i$ to $j$,what can we say about $Iso(X,d)$ ? Are there any books about this literature? Thanks.
As secretman says, the condition that there exist $i$ and $j$ in $X$ such that no isometry carries $i$ to $j$ is precisely to say that the action of the isometry group on $X$ is not transitive. If that's really your question, that seems to be all that can be said, and I doubt there any books on the subject. One might simply ask: what can be said about isometry groups of finite metric spaces? This is a rather broad question, but nevertheless it seems interesting. Here is what I was able to come up with via a short amount of thought and googling: 1) Every finite group $G$ occurs up to abstract group isomorphism as the full isometry group of a finite metric space. Indeed, in this 1976 paper, D. Asimov proved that if $G$ is finite of cardinality $k$, there exists a finite subset $X_G$ of Euclidean $k1$space (with the induced metric), of cardinality $k^2k$, such that the full isometry group of $X_G$ is isomorphic to $G$. 2) On the other hand, it is not true that every permutation group is the isometry group of a finite space: i.e., if $X = \{1,\ldots,n\}$ and $G$ is subgroup of $S_n$, then there need not be a metric $\rho$ on $X$ such that $G = \operatorname{Aut}(X,\rho)$. For a simple example, take $n = 3$ and let $G$ be the subgroup of order $3$, i.e., $G$ is generated by $\sigma = (123)$. Since $\sigma$ is an isometry, $\rho(1,2) = \rho(\sigma(1),\sigma(2)) = \rho(2,3) = \rho(\sigma(2),\sigma(3)) = \rho(1,3)$. Thus all pairwise distances between the three elements of $X$ are equal and the isometry group is the full $S_3$. More generally, let $\rho$ be a metric on $X$ and $G \subset \operatorname{Sym}(X)$ the isometry group. Let $X^{(2)}$ be the set of (unordered!) twoelement subsets of $X$: there is a natural action of $\operatorname{Sym}(X)$  and hence also of $G$  on $X^{(2)}$. Now the observation here is that if $G$ acts transitively on $X^{(2)}$  e.g. if $G$ is doubly transitive as a permutation group on $X$, but this condition is weaker  then that means that all pairs of points in $X$ have the same distance, and therefore $G = \operatorname{Sym}(X)$.



In accordance with Huichi Huang's comment on secretman's answer, we can say the following: If no isometry sends $i$ to $j$, then there exists some $k$ (Edit: different from $i$ and $j$) with $d(i,k)\neq d(j,k)$. Why? Because if $d(i,k)=d(j,k)$ for all $k$, then the transposition $(ij)\in S_n$ is an isometry of $X$. 

