Yes, if a Hilbert space $H$ is infinite dimensional, then $H\otimes H$ is isomorphic to $H$.
Recall that Hilbert spaces are completely determined by the cardinality of their orthonormal basis up to isomorphism, i.e. if a Hilbert space $K$ has an orthonormal basis $\{e_i\}_{i\in I}$, then $K\cong \ell^2(I)$, where $\ell^2(I)$ is the space of those $f\colon I\to\mathbb{K}$ with $\sum_{i\in I}|f(i)|^2<\infty$.
Moreover, if $I,J$ are two sets and $\phi:I\to J$ is a bijection, then we get an induced unitary operator $U:\ell^2(I)\to\ell^2(J)$ satisfying $U\delta_i:=\delta_{\phi(i)}$ for all $i\in I$, where $\{\delta_i\}_{i\in I}$ is the canonical orthonormal basis of $\ell^2(I)$, namely $\delta_i(i')=1$ for $i'=i$ and $\delta_i(i')=0$ for $i'\ne i$.
Now if $H$ is infinite dimensional and $\{e_i\}_{i\in I}$ is an orthonormal basis for $H$, then $\{e_i\otimes e_j\}_{(i,j)\in I\times I}$ is an orthonormal basis for $H\otimes H$. As $I\times I$ and $I$ have the same cardinality (for which we assume the axiom of choice -- see here), we see that $H\otimes H\cong H$.
As for how $D\otimes T$ acts on $H$, this depends on the choice of unitary operator between $H\otimes H$ and $H$ that you choose; these are in a correspondence with the bijections between $I\times I$ and $I$, which are of course many.