If liquid hydrogen fluoride autoionises, how can it be an insulator?
Unlike metals which can conduct with free electrons, conduction in non-metallic liquids (ionic and covalent) is dependent on the ion concentration to carry charges. Pure water has a autoionization constant of $\rm K_w = 1\!\times\!10^{-14}$, which corresponds to a total ion concentration of $\rm 2\!\times\!10^{-7} \pu M$. At this concentration the resistivity of water is $\pu{18.2 M\Omega~ cm}$ which corresponds to a conductivity of $\pu{5.5 \mu S/cm}$ and is considered to be a poor conductor. Sea water on the other hand has about $\pu{3.4\%}$ sodium chloride and a conductivity of about $\pu{55,000 \mu S /cm}$ which is a million times improvement in conductivity which is considered conductive.

Hydrogen fluoride has an autoionization constant of $\rm 10^{-12.5} ~~(3.162\!\times\!10^-13)$ which corresponds to a total ion concentration of $\rm 1.125\!\times\!10^{-6}$ which is about five-times that of water. Five times the ions gives us five times the conductivity of $\pu{275\mu S/cm}$ or a resitivity of $\pu{2.28M\Omega ~cm}$ which is still a quite poor conductor. Thus it is necessary to add an ionic substance to increase conductivity to have appreciable conduction of the hydrogen fluoride similar to sea water.