= Solution
Let $V=\mathbb F_2^3$. Since $\mathbb F_2^\times=\{1\}$ and the <scalar> centre is trivial,
$$
G=\mathrm{PSL}(3,2)=\mathrm{GL}(3,2),\qquad |G|=(8-1)(8-2)(8-4)=168.
$$
Choose $U_1$ to stabilize a nonzero <vector> and $U_2$ to stabilize a two-dimensional subspace. There are seven choices of either object, so both <subgroups> have index $7$ and order $24$.
Their coset <permutation characters> agree. The number of nonzero <vectors> fixed by $g$ is $2^{\dim\ker(g-I)}-1$. A two-dimensional subspace is the kernel of a unique nonzero <covector>, so its stabilizer action is the dual action $g^{-T}$. Its fixed-point count is
$$
2^{\dim\ker(g^{-T}-I)}-1=2^{\dim\ker(g-I)}-1,
$$
using equality of the <matrix ranks> of a <matrix> and its transpose. For a <subgroup> $H$, counting the representatives of fixed cosets gives
$$
\chi_{G/H}(g)=\frac{|C_G(g)|}{|H|}\,|H\cap[g]_G|.
$$
Thus the equal characters and <subgroup> orders give equal intersection counts with every <conjugacy class>. This proves \b[$U_1,U_2$ are <Gassmann equivalent>].
They are not conjugate. For example, the stabilizer of $e_1$ contains every <matrix> $\begin{pmatrix}1&b\\0&B\end{pmatrix}$ with $b\in\mathbb F_2^{1\times2}$ and $B\in\mathrm{GL}(2,2)$. An invariant plane containing $e_1$ would give an invariant line in $V/\langle e_1\rangle$, impossible for the full $\mathrm{GL}(2,2)$ action. An invariant plane not containing $e_1$ would be a complement to that line, but the arbitrary shears $b$ do not preserve any such complement. Hence a point stabilizer preserves no plane, whereas every conjugate of $U_2$ does. \b[The two <subgroups> are nonconjugate.]
We now use a <cone-torus construction of genus-four Sunada surfaces>. Let $B$ be a hyperbolic torus with one cone point of angle $2\pi/7$. Its <orbifold fundamental group> has presentation
$$
\Gamma=\langle a,d\mid[d,a]^7=1\rangle.
$$
Map $a,d$ to the supplied generators $A,D$ of $G$. Their commutator has exact order $7$, so this gives a surjection $\Gamma\to G$ with torsion-free kernel $K$: every finite-order element of $\Gamma$ is conjugate to a power of the cone generator, and no nonidentity such power lies in $K$. Thus $N=\mathbb H^2/K$ is a smooth closed <hyperbolic surface> with a $G$-action. Its quotients by $U_1,U_2$ are smooth too, since order-$24$ <subgroups> contain no element of order $7$ and therefore meet no cone stabilizer.
The <orbifold Euler characteristic> of $B$ is $-(1-1/7)=-6/7$. Each quotient cover has degree $7$, so
$$
\chi(S_i)=7\chi_{\rm orb}(B)=-6=2-2g(S_i),\qquad
\boxed{g(S_1)=g(S_2)=4}.
$$
The <Sunada theorem> proved in Question 3 makes $S_1,S_2$ isospectral for every choice of the base <hyperbolic metric>.
Here is an explicit supply of base metrics. A <hyperbolic trirectangle> has three right angles and a fourth angle $\alpha=\pi/14$. If its sides adjacent to the opposite right-angle vertex have lengths $r,s$, the trirectangle identity is $\sinh r\sinh s=\cos\alpha$. Thus $r>0$ varies freely, with $s=\operatorname{arsinh}(\cos\alpha/\sinh r)$. Reflecting four copies about their two perpendicular centre lines gives a quadrilateral with four angles $\pi/14$ and opposite sides of equal length. Identifying opposite sides produces a cone torus: all four corners give total angle $2\pi/7$. Its two centre-line <geodesics> intersect once and have lengths $2r,2s$. Cut along the shorter chosen <geodesic> $d$ and reglue with a small twist $\tau$; this retains the cone angle and allows an additional metric parameter. The remaining cut surface has two equal-length <geodesic> boundaries and the cone point.
We give a concrete <intersection test for nonisometric finite covers>, rather than relying on <subgroup> nonconjugacy alone. Choose $\ell(d)=\varepsilon$ sufficiently small. The trirectangle geometry gives an embedded collar of width $\log(1/\varepsilon)+O(1)$ about $d$. The shortest perpendicular joining the two boundary components after cutting has length $\delta=2\log(1/\varepsilon)+O(1)$. At zero twist it closes to $a$, the unique shortest <closed geodesic> crossing $d$ once. Uniqueness follows from uniqueness of the shortest perpendicular; all other once-crossing classes have a positive length gap, since only finitely many <closed geodesics> have bounded length. For sufficiently small twists, the same marked class $a$ remains the unique shortest once-crossing <geodesic>. A <geodesic> crossing $d$ at least twice traverses the collar at least twice, so for small $\varepsilon$ its length is greater than $\ell(a)$.
<Geodesics> disjoint from $d$ remain entirely in the cut surface and have twist-independent lengths. In a bounded window around $\ell(a)$ they supply only finitely many possible length values, including lengths of iterates. Meanwhile $\ell(a)$ varies nontrivially with $\tau$ by the hyperbolic seam identity: its hyperbolic cosine has a positive multiple of $\cosh(\tau/2)$. Choose a small twist away from those finitely many coincidences. Then the only base <geodesic> of length $\ell(a)$ is $a$, up to <orientation>. By taking $\varepsilon$ smaller if necessary, the only base <geodesics> of length at most $2\varepsilon$ are powers of $d$: those crossing $d$ are long, and the nonperipheral <geodesics> of the limiting two-cusped cone pair of pants have lengths bounded away from zero.
In each cover, primitive lift components of a base <geodesic> are indexed by cycles of its <monodromy permutation>, with length equal to the cycle size times its base length. The permutation of $A$ has a unique singleton $\{0\}$ in each action, so each $S_i$ has a uniquely length-identified primitive lift $\alpha_i$ of length $\ell(a)$. The permutation of $D$ has one two-cycle: it is $\{0,3\}$ for $U_1$ and $\{2,5\}$ for $U_2$. Hence each $S_i$ has a unique primitive lift $\beta_i$ of length $2\varepsilon$; the iterate of the singleton lift is not primitive. At the single base intersection of $a,d$, intersections of their lifted components are exactly their common sheet labels. Therefore
$$
\boxed{i(\alpha_1,\beta_1)=1,\qquad i(\alpha_2,\beta_2)=0}.
$$
An <isometry> would have to preserve the uniquely identified primitive lengths and their <geometric intersection numbers>. This contradiction proves that $S_1$ and $S_2$ are not isometric, even allowing <orientation> reversal.
Finally, let $\varepsilon$ range over a sufficiently small positive interval, choosing a permissible twist for each value. The singleton cycle of $D$ makes the <hyperbolic systole> of each cover equal to $\varepsilon$. Distinct values therefore give distinct <isometry> classes. We obtain \b[uncountably many pairs of nonisometric, isospectral <genus>-four <Riemann surfaces>].
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