Are there examples of (distinct) Hilbert spaces $H\_1$=$(H,\langle\cdot,\cdot\rangle\_1)$, $H\_2 $=$(H,\langle\cdot,\cdot\rangle\_2)$ and a linear operator $V: H\_1\to H\_2$ such that $V^n: H\_1\to H\_2$ is an isometry $\mathbf{??}$ for every $n=1,2,...$.


Note that: 


1) $H\_1$ and $H\_2$ have the same underlying linear space structure, and we look for $V$: $||V^n(x)||\_2=||x||\_1$ for all $x\in H\_1$ with $||x||\_k=\sqrt{\langle x,x\rangle\_k}$, $k=1,2$)


2) if $V: K\to K$ is an isometry on a Hilbert space $K$, so is $V^n$ (i.e. $||V^n(x)||=||x||$, for all $x\in K$), so the question is if we can find such isometries between $\underline{\rm distint}$ Hilbert space structures. 



Examples of special isometries
The answer depends on whether you ask $V$ to be invertible (you do not seem to do, but the usual definition of an isometry includes it).


(**Edit:** paragraph corrected thanks to Nick Weaver's comment)
If $V$ is not asked to be invertible, then the answer is **yes**: take $H=\ell^2(\mathbb{N})$, $\langle\cdot,\cdot\rangle\_1$ be the usual $\ell^2$ product, $V$ be the right shift and $\langle\cdot,\cdot\rangle\_2$ be equal to $\langle\cdot,\cdot\rangle\_1$ except that it is rescaled on the first coordinate (e.g. $\langle u,v\rangle\_2=2u\_1v\_1+\sum\_{i>1} u\_iv\_i$).


If $V$ is asked to be invertible, then the answer is **no**: simply look at the action of $V$ on scalar product 
$$V\cdot\langle\cdot,\cdot\rangle = \langle V^{-1}\cdot,V^{-1}\cdot\rangle.$$
Then you ask for two distinct points (in the space of scalar products) such that one is fixed and the other is mapped to the first one. This is clearly impossible for an invertible map.


