Corollary to Putnam's theorem Suppose $T_1$ and $T_2$ are normal operators on Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$, respectively. Putnam showed that if $X$ is an operator satisfying $T_2X=XT_1$, then $T_2^*X=XT_1^*$. 
Suppose $T_1$ and $T_2$ are similar, so that there is some operator $S$ with a two-sided inverse such that $ST_1=T_2S$. How can one show that $T_1$ and $T_2$ are unitarily equivalent?
This result is mentioned in a number of papers, but I can't find a proof. There is a purported proof in the Wikipedia article on Fuglede's theorem (where a correct proof of Putnam's result can be found), but it is obviously erroneous. 
 A: It is fairly straightforward to reduce it to the case $\mathcal{H}_1 = \mathcal{H}_2$ treated in Rudin's theorem 12.36.
Consider the Hilbert space $\mathcal{H} = \mathcal{H}_1 \times \mathcal{H}_2$. A linear operator on $\mathcal{H}$ decomposes
$$L = \begin{pmatrix}\alpha & \beta\\ \gamma & \delta\end{pmatrix},$$
with $\alpha \colon \mathcal{H}_1 \to \mathcal{H}_2$, $\beta \colon \mathcal{H}_2 \to \mathcal{H}_1$, $\gamma\colon \mathcal{H}_1 \to \mathcal{H}_2$ and $\delta \colon \mathcal{H}_2 \to \mathcal{H}_2$. We then have
$$\begin{pmatrix}\alpha & \beta\\ \gamma & \delta\end{pmatrix}^\ast  = \begin{pmatrix}\alpha^\ast & \gamma^\ast\\ \beta^\ast & \delta^\ast\end{pmatrix}.$$
The operators
$$N_1 = \begin{pmatrix} T_1 & 0\\0&0\end{pmatrix};\quad N_2 = \begin{pmatrix}0&0\\ 0&T_2 \end{pmatrix}$$
are then normal, and the invertible operator
$$B = \begin{pmatrix}0&S^\ast\\ S&0\end{pmatrix}$$
satisfies $BN_1 = N_2 B$. Let $S = UP$ be the polar decomposition of $S$. The polar decomposition of $B$ is then
$$B = \begin{pmatrix}0&U^\ast\\ U&0\end{pmatrix} \begin{pmatrix}P&0\\ 0&UPU^\ast\end{pmatrix} = \mathscr{U}\mathscr{P}.$$
By Rudin's theorem 12.36, we have $\mathscr{U}N_1 = N_2 \mathscr{U}$, and that translates to $UT_1 = T_2 U$.

Of course, shortly after I turned the computer off, I saw an even simpler way to reduce the problem to theorem 12.36.
Let $S = UP$ the polar decomposition of $S$, and define $T_3 = U^\ast T U$. Then $T_3 \colon \mathcal{H}_1 \to \mathcal{H}_1$ is normal, and from $ST_1 = T_2 S$ we obtain $U^\ast ST_1 = U^\ast T_2 S = (U^\ast T_2 U)(U^\ast S) = T_3 (U^\ast S)$. The polar decomposition of $U^\ast S = U^\ast U P$ is $U^\ast S = (U^\ast U) P = I\cdot P$, so by 12.36, we have $I\cdot T_1 = T_3\cdot I$, i.e. $T_1 = U^\ast T_2 U \iff UT_1 = T_2 U$.
