Show that if x and y are of same length, there must exist an isometry with f(x) = y

Question

Assume $x, y \in V$ with $V$ being an euclidean vector space. I have to show:

There exists an isometry $f: V \rightarrow V$ with $f(x) = y$, if and only if $\left|\left|x\right|\right| = \left|\left|y\right|\right|$.

I already showed the direction "$\Rightarrow$", but "$\Leftarrow$" is giving me problems. I know that for every $x,y \in V$ I can just pick a linear mapping $f : V \rightarrow V$ with $f(x) = y$ and show that it is an isometry using the fact that $\left|\left|x\right|\right| = \left|\left|y\right|\right|$, but I just can't seem to find a good pick!

Answer 1

Let $M$ be the two dimensional subspace spanned by $x$ and $y$. If $N$ is $M^{\perp}$ then $V=M+ N$. If you can find an isometry $S$ on $M$ such that $Sx=y$ then you can get an isometry $T$ on $V$ by defining $T(u+v)=Su+v$ for $u \in M,v \in N$. Hence the question reduces to two dimensional Euclidean space. Here there is a rotation which takes $x$ to $y$.