If a non-linear map preserves inner product, does it necessarily preserve distance?

Question

Let $(S, \langle\, , \rangle)$ be a prehilbertian space.

I know that if a map $f: S \rightarrow S$ preserves norm but is not linear then it need not preserve metric. For instance, consider $f : (r\cos (\theta ), r\sin (\theta )) \mapsto$ $(r\cos (r\theta ), r\sin (r\theta )) $. We have $d(f(0,1),f(0,2))=||f(0,1)-f(0,2)||=||(0,1)-f(0,2)|| > 1=d((0,1),(0,2))$

Now I'm trying to find a counterexample for a map that preserves inner-product but, as above, doesn't preserve metric, but I've got a feeling this doesn't exist, am I right ?

Answer 1

Assume that the inner product space is real. (One can do the same computation for a complex inner product space.) Note that $$ (f(x),f(y))=(x,y),\quad\forall x,y\in S $$ implies that $$ \|f(x)-f(y)\|^2=(f(x)-f(y),f(x)-f(y))=(x,x)+(y,y)-2(x,y)=\|x-y\|^2. $$ and thus the distance must be preserved.