When bosonizing an interacting spinless Luttinger liquid, the action can be written as \begin{equation} S=\frac{K}{2\pi}\int dx d\tau\ (\partial_\mu\phi)^2 = \frac{1}{2\pi K}\int dx d\tau\ (\partial_\mu\theta)^2, \end{equation} where $$K=\sqrt{\frac{v_F+g_4/\pi+g_2/\pi}{v_F+g_4/\pi-g_2/\pi}}$$ is the Luttinger parameter, which is one for free fermions. The convention for the $\phi$ field is such that its compactification radius is $R=1$. Alternatively $K$ can be absorbed into the definition of the fields to change $R$ to $R=\sqrt{K}$.
The first equation has an apparent symmetry under $K\to 1/K$ and $\theta\to\phi$, and the free fermion case lies right on the self-dual point $K=1$ ($R=1$).
This duality looks very similar to the $T$-duality for a compact free boson CFT. In fact on page 157 of Fradkin's book (pdf available online), it was explicitly pointed out that "in string theory this transformation is known as T-duality and the Luttinger parameter is known as the compactification radius (see e.g. Polchinski (1998) and Di Francesco et al. (1997))."
However, in CFT it is well known that the $T$-duality takes $R\to 1/(2R)$ (or $K\to 1/(4K)$ using the convention above) and the self-dual point is $R^*=1/\sqrt{2}$, rather than $R=1$. Moreover, assuring this is not just a naive convention issue, there is an emergent $SU(2)\times SU(2)$ symmetry at this self-dual radius $R^*$, which is not the case for a free fermion theory.
So I am really confused whether the $K=1$ case is self-dual under $\theta\to\phi$, and if it is, whether it has anything to do with the $T$-duality. Is the statement in Fradkin's book quoted above wrong? What is the relation between this "$\theta$-$\phi$ duality" and $T$-duality?