Let $\mathcal{H}$ be a Hilbert space, $A, B\in B(\mathcal{H})$ be two normal operators. Suppose $T$ is an invertible operator of $B(\mathcal{H})$ such that $TA=BT$. How to show that there is an unitary operator $U\in B(\mathcal{H})$ such that $UA=BU$?
Since $A$ and $B$ are both normal, $A$ has a unique decomposition $A=U_1 P_1 = P_1 U_1$, $B$ has a unique decomposition $B=U_2 P_2 = P_2 U_2$, where $U_1$, $U_2$ are unitary, $P_1$, $P_2$ are self-adjoint. Then we have $$ TU_1 P_1 = U_2 P_2 T $$ Since $T$ is invertible ,there is an operator $S$ such that $TS=ST=I$. So $U_1 P_1 = S U_2 P_2 T$.
Then what to do next? Thanks in advance!
It turns out that this is a well-known(?) application of the Fuglede-Putnam Theorem (see Conway for a slick proof). This Wikipedia article has a proof of the result you seek. I will write it down here for future readers.
So by the Fuglede-Putnam Theorem, $TA^*=B^*T$. Since $T$ is invertible, we get $B^*=TA^*T^{-1}$. Taking adjoints on both sides, $$(T^{-1})^*AT^*=B=TAT^{-1}.$$ Thus, we get $$(T^*T)^{-1}A(T^*T)=A.$$ Since $T$ is invertible, we can write $T=U|T|$ with $U$ unitary and $|T|=(T^*T)^{-1/2}$. Since $(T^*T)$ commutes with $A$, $|T|$ also commutes with $A$ ($|T|$ is a limit of polynomials in $(T^*T)$). Furthermore, $|T|$ is invertible. Thus $$|T|A|T|^{-1}=A.$$ Finally, we have$$B=TAT^{-1}=U|T|A|T|^{-1}U^{-1}=UAU^{-1}.$$