Equivalence of norms implies isomorphism between Hilbert spaces If I have 2 Hilbert spaces with 2 norms, and a map between the Hilbert spaces, and I know that the norms are equivalent, does this mean that the spaces are isomorphic? 

I assume that the mentioned map between the Hilbert spaces (say, $\mathcal H_1$ and  $\mathcal H_2$) is a linear bijection, say $\varphi:\mathcal H_1\to\mathcal H_2$, such that the norms are equivalent w.r.t. this $\varphi$, i.e.
$$ \exists c,C>0: \ \forall x\in\mathcal H_1: \  
c||x||_1 \le ||\varphi(x)||_2 \le C||x||_1 $$
I think we also need to assume separability, at least, it would need more work for the general case.
Let us fix an orthonormal basis $(b_i)$ for $\mathcal H_1$, and consider a mapping
$\psi:\mathcal H_1\to\mathcal H_2$ obtained via the Gram-Schmidt orthogonalization project from $\varphi(b_1),\varphi(b_2),\dots$, yielding an orthonormal basis $\psi(b_1)$ starting as
$$\psi(b_1):=\frac{\varphi(b_1)}{||\varphi(b_1)||},\ 
\psi_0(b_2):=\varphi(b_2)-\langle \varphi(b_2),\psi(b_1)\rangle\cdot \psi(b_1) \\
\psi(b_2):=\frac{\psi_0(b_2)}{||\psi_0(b_2)||},\ \ldots
$$
Using that $\varphi$ is bounded. This way, $\psi$ preserves the norm and orthogonality of the basis vectors, hence, is an isomorphism.
