The following is Exercise II.8.7 from J. B. Conway's A Course in Functional Analysis.
Let $A:L^2(0,1)\to L^2(0,1)$ be defined by $(Af)(x)=xf(x)$ for $f$ in $L^2(0,1)$ and $x$ in $(0,1)$. Show that $A\cong A^2$.
Here $A\cong A^2$ means there is an unitary isomorphism $U:L^2(0,1)\to L^2(0,1)$ such that $UAU^{-1}=A^2$. I guess I should construct an explicit isomorphism, because $A$ is not compact, and the author has not proved any structure theorem for non-compact operators yet. However, I don't know how I can find such an isomorphism. I understand that $L^2(0,1)$ has Hilbert basis $e^{2\pi inx}$, and I computed the matrix of $A$ with respect to this basis, but I can't see how $A\cong A^2$.
Any hints will be appreciated!
Let $V\in B(L^2(0,1))$ be given by $$ Vf\,(x)=\frac{f(\sqrt x)}{\sqrt2\,x^{1/4}}. $$ This map is invertible: if $$ Wg\,(x)=\sqrt{2x}\,g(x^2) $$ then $VW=WV=I$, so $V$ is invertible. Also, $$\tag1 \|Vf\|^2=\int_0^1\frac{|f(\sqrt x)|^2}{2\,x^{1/2}}\,dx=\int_0^1|f(u)|^2\,du=\|f\|^2. $$ So $V$ is a unitary. And, $$ VA^2f\,(x)=V(x^2f)\,(x)=\frac{x\,f(\sqrt x)}{\sqrt2\,x^{1/4}}=AVf\,(x). $$ This gives us $VA^2=AV$, and then $VA^2V^*=A$.