We call a map between 2 normed vector spaces is isometric isomorphism if it is an isometry and an linear isomorphism. Then, these 2 vector spaces are called isometrically isomorphic.
Suppose $X,\,Y$ be 2 metric vector space of dimension $n$. Suppose $f:\ X\longrightarrow Y$ is a isometric, that is $$d_X(x,y)=d_Y\big(f(x),f(y)\big),\ \forall x,y\in X.$$ I want to know if $X,Y$ are linear isomorphic ?
What I know is if $X,Y$ are normed $n-$vector space, then every isometry $f:\ X\longrightarrow Y$ must be affine, which means \begin{align} f(x)=L(x)+u, \end{align} for some bijective linear $L:\ X\longrightarrow Y$ and $u\in Y$. Then $L$ will be a isometric isomorphism.
But is it true for the preceding case ? Moreover, if 2 metric vector spaces $n$ dimensional are isometric, then would they be isomorphic ?
I hope someone will help me to clarify this. Thanks
The result that you have cited about isometries between normed spaces being affine (which is linear plus a constant and subtracting the constant gives a linear isometry) is that the norm, by its definition, has some interplay with the linear structure. For example, homogeneity: $\|tx\|=|t|\|x\|$ connects the norm to scalar multiples (algebraic operation). The triangle inequality connects the norm to sums (algebraic operation).
If we drop this and ask about general metrics which don't have any particular connection to the linear structure of the space, we don't get any nice result. For example, take $X=Y=\mathbb{R}$. Define $$d_X(x,y)=\left\{\begin{array}{ll} 0 & : x=y \\ 1 & : x\neq y, \{x,y\}\cap \{0\}=\varnothing,\\ 2 & : x\neq y, \{0\}\subsetneq \{x,y\}.\end{array}\right.$$
So any two non-zero points are distance $1$ from each other, and $0$ is distance $2$ from everything else.
$$d_Y(x,y)=\left\{\begin{array}{ll} 0 & : x=y \\ 1 & : x\neq y, \{x,y\}\cap \{1\}=\varnothing,\\ 2 & : x\neq y, \{1\}\subsetneq \{x,y\}.\end{array}\right.$$
Here, any two non-$1$ points are distance $1$ from each other, and $1$ is distance $2$ from everything else.
There are lots of isometries from $X$ to $Y$ (any bijection $F:X\to Y$ which has $F(0)=1$ is an isometry). But there are no linear isometries, because a linear operator takes $0$ to $0$ and an isometry from $X$ to $Y$ has to take $0$ to $1$.