It is said that if $X$ is a pre-Hilbert space(inner product space), then for any $x, y\in S_X$(unit sphere of $X$), there is a surjective linear isometry $T$ from $X$ to $X$ such that $Tx=y$.
I cannot actually find a map from $X$ to $X$. Is there a map satisfies the condition? or another way to prove it without find the map?
One can employ the two-dimensional case to prove the general result: assume that $x \neq \pm y$ (else it's obvious what to do). Then $\{x,y\}$ spans a two-dimensional subspace, and it suffices to define a linear isometry on this subspace that sends $x$ to $y$, and extend it linearly by the identity.
This works because one can orthogonally decompose a vector in $X$ uniquely into $u+v$ where $u \in U=\operatorname{span}\{x,y\}$ and $\langle u, v \rangle = 0$ (by using projection, essentially). Then $Tv=v$ and $Tu \in U$, so $\langle Tu,Tv \rangle = 0$, and since $T$ is an isometry on $U$, $$ \langle T(u+v),T(u+v) \rangle = \langle Tu+v,Tu+v \rangle = \langle Tu,Tu \rangle + \langle v,v\rangle = \langle u,u\rangle +\langle v,v\rangle = \langle u+v,u+v \rangle, $$ so $T$ is an isometry of $X$ as well.