= Solution
For each fixed $k$, the two quotient <Riemannian metrics> are not isometric. If an <isometry> $F:N_1\to N_2$ existed, choose a small open set avoiding the finite preimages of cone points and their inverse images under $F$. Composing $F$ with the local inverse of the first base projection and the second base projection gives a <local isometry> of the base <Riemannian metric> $h_k$. Local rigidity forces this map to be the identity. Hence $p_2F=p_1$ on a dense open set, and by <continuity> everywhere.
The alleged <isometry> would therefore be an isomorphism of the two <orbifold> covers over the same base. Covering-space classification makes $K_1,K_2$ conjugate in $\Gamma$, and applying $\rho$ makes $H_1,H_2$ conjugate in $T$, contrary to the supplied hypothesis. Thus $g_{1k},g_{2k}$ are nonisometric.
For $k\ne\ell$, the surface areas are different:
$$
\boxed{\operatorname{area}(N,g_{ik})=4\pi(1+2^{-k}).}
$$
Area is preserved by <isometries>, so no surface from the $k$th pair is isometric to a surface from the $\ell$th pair. Together these arguments prove the full pairwise nonisometry assertion.
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