Isometries of direct sums of Hilbert spaces If $H$ is a separable Hilbert space, then for any norm one vectors $x$ and $y$ we can find an surjective isometry $U$ such that $Ux=y$. Is the same true for $H\oplus H$, that is, the direct sum in $l_1$ sense? This space is isomorphic, but not isometric with $H$, so I am not sure the property still holds. Of course, if we consider the direct sum $\oplus_2$ in $l_2$ sense, then the space is isometric to $H$ and the property again holds.
I suspect it is not true, but I cannot come up with a pair of points for which it fails.
 A: Let $V$ be an isometry in your sense. Being a linear operator on $H\oplus H$, we can think of $V$ as a $2\times 2$ matrix of operators. That is,
$$
V=\begin{bmatrix} A&B\\ C&D\end{bmatrix},
$$
with $A,B,C,D\in B(H)$. That is,
$$
V(x,y)=(Ax+By,Cx+Dy),\qquad\qquad x,y\in H. 
$$
Note that since $\|(x,y)\|_2\leq\|(x,y)\|_1\leq\sqrt2\,\|(x,y)\|_2$, the bounded operators on $H\oplus H$ are the same regardless of the norm.
Since $V$ is an isometry, we have
$$
\|x\|+\|y\|=V(x,y)=\|(Ax+By,Cx+Dy)\|=\|Ax+By\|+\|Cx+Dy\|. 
$$
Taking $y=0$,
$$\tag1
\|x\|=\|Ax\|+\|Cx\|,\qquad\qquad x\in H.
$$
Taking $x=0$,
$$\tag2
\|y\|=\|By\|+\|Dy\|,\qquad\qquad y\in H.
$$
Taking $y=\lambda x$ with $\|x\|=1$, and $|\lambda|=1$,
$$\tag3
2=\|(A+\lambda B)x\|+\|(C+\lambda D)x\|,\qquad\qquad x\in H.
$$
Combining $(1)$, $(2)$, and $(3)$, for $x\in H$ with $\|x\|=1$ we have
$$\tag4
\|Ax\|+\|Bx\|+\|Cx\|+\|Dx\|\leq2=\|(A+\lambda B)x\|+\|(C+\lambda D)x\|.
$$
This implies equality in both triangle inequalities, so for any $x\in H$ and $|\lambda|=1$
$$\tag5
\|Ax\|+\|Bx\|=\|(A+\lambda B)x\|,\qquad \|Cx\|+\|Dx\|=\|(C+\lambda D)x\|. 
$$
We now work with $A,B$ since the computations for $C,D$ are entirely analogous. Squaring, expanding, and cancelling square norms in $(5)$,
$$\tag6
\operatorname{Re}\lambda \langle Bx,Ax\rangle=\|Bx\|\,\|Ax\|,\qquad\qquad x\in H,\ |\lambda|=1. 
$$
By using $\lambda=1,-1,i,-i$ we get that
$$\tag7
\|Bx\|\,\|Ax\|=0,\qquad\qquad x\in H.
$$
and analogously
$$\tag8
\|Cx\|\,\|Dx\|=0,\qquad\qquad x\in H.
$$
When $Ax=0$, we get from $(1)$ that $\|Cx\|=\|x\|$; then $(8)$ implies that $Dx=0$. Similarly,  when $Bx=0$ we get that $\|Dx\|=\|x\|$ and then $Cx=0$. So either
$$\tag9
\|Ax\|=\|Dx\|=0,\qquad\qquad \|Cx\|=\|Bx\|=\|x\|,
$$
or
$$\tag{10}
\|Ax\|=\|Dx\|=\|x\|,\qquad\qquad \|Cx\|=\|Bx\|=0.
$$
Suppose that $(9)$ occurs for a certain $x$ and $(10)$ for $y$, both nonzero. For $x+y$, we have
$$
\|A(x+y)\|=\|Ay\|=\|y\|,\qquad \|B(x+y)\|=\|Bx\|=\|x\|,
$$
so $x+y$ satisfies neither $(9)$ nor $(10)$. This proves that $V$ satisfies either $(9)$ or $(10)$ for all $x$. In other words, the possibilities for $V$ are
$$
V=\begin{bmatrix} A&0\\0&D\end{bmatrix},\qquad\qquad\text{ or } \qquad\qquad V=\begin{bmatrix}0& B\\ C&0\end{bmatrix}
$$
with $A,B,C,D\in B(H)$ isometries.
Now it is easy to find the counterexample. Fix $x\in H$ with $\|x\|=1$ and consider the elements $(x,0)$ and $(x/2,x/2)$. With $V$ of the first form we need to have
$$
(x/2,x/2)=V(x,0)=(Ax,0),
$$
and this forces $x=0$. With $V$ of the second form the problem is the same:
$$
(x/2,x/2)=V(x,0)=(0,Cx)
$$
and we get $x=0$. So no isometry (surjective or not) can map $(x,0)$ to $(x/2,x/2)$.
