= Solution
The <Sunada theorem> is the following. Let $X\to M_0$ be a finite normal <Riemannian covering> of a connected closed <Riemannian manifold>, with finite <deck transformation group> $T$. If $U_1,U_2\leq T$ satisfy <Gassmann equivalence>, meaning
$$
|U_1\cap C|=|U_2\cap C|\quad\text{for every conjugacy class }C\text{ of }T,
$$
then $U_1\backslash X$ and $U_2\backslash X$ have the same scalar <Laplacian spectrum>, with multiplicities. Equivalently, an epimorphism $\pi_1(M_0)\to T$ yields isospectral covers corresponding to the inverse images of $U_1,U_2$.
Here is the representation-theoretic proof. Every finite-dimensional <Laplacian eigenspace> $E_\lambda$ of $X$ is a <group representation> of $T$, since <isometries> commute with the <Laplace-Beltrami operator>. Eigenfunctions on $U_i\backslash X$ identify with $E_\lambda^{U_i}$. Averaging gives the projection $P_i=|U_i|^{-1}\sum_{u\in U_i}u$, so
$$
\dim E_\lambda^{U_i}=\frac1{|U_i|}\sum_{u\in U_i}\chi_\lambda(u).
$$
The two subgroup orders agree, and the <character of a representation> $\chi_\lambda$ is constant on <conjugacy classes>. The displayed multiplicities are therefore equal for every $\lambda$. This proof also applies when $T$ has fixed points on $X$, provided the two subgroups $U_i$ act freely, so their quotients are smooth <manifolds>. We shall use this extension for an orbifold base in Question 4.
The precise geometric test for an <isometry> uses the full isometry group of a common <universal cover>. Equip the simply connected cover $\widetilde X$ of $M_0$ with the lifted metric and let $\Gamma_i$ be the subgroups of its <deck transformation group> corresponding to the two covers. Then
$$
\boxed{\Gamma_1\backslash\widetilde X\cong\Gamma_2\backslash\widetilde X\text{ isometrically}\iff
F\Gamma_1F^{-1}=\Gamma_2\text{ for some }F\in\operatorname{Isom}(\widetilde X).}
$$
Indeed, such an $F$ descends to an <isometry> of the quotients. Conversely an <isometry> of the quotients lifts to their simply connected covers; lift its inverse with compatible basepoints to see that the lift is bijective. Since it preserves the lifted metrics locally, it is a global <isometry>, and equivariance of the covering maps shows that it conjugates the deck subgroups. This proves the criterion, rather than assuming that an isometry must preserve the map to $M_0$. In particular, nonconjugacy inside $T$ alone does not establish nonisometry; the full covering space may have additional isometries. Nonisomorphic <fundamental groups> are a sufficient topological obstruction, and length-identified geometric configurations provide another.
For a concrete <nonisomorphic Gassmann equivalent regular subgroups> example, fix an odd prime $p$ and take $T=S_{p^3}$. Embed $E=(C_p)^3$ and $B=\operatorname{UT}_3(\mathbb F_p)$ through their regular permutation actions. Both have order $p^3$. Every nonidentity element of $E$ has order $p$. For $B$, write its elements as $I+N$ with $N^3=0$; then
$$
(I+N)^p=I+pN+\binom p2N^2=I
$$
over $\mathbb F_p$, including $p=3$, so its nonidentity elements also have order $p$. In a regular action, each nonidentity order-$p$ element has cycle type $p^{p^2}$. Thus each subgroup meets the identity class in one element, the class of that cycle type in $p^3-1$ elements, and every other <conjugacy class> in zero elements. They are <Gassmann equivalent>. They are not isomorphic: $E$ is abelian, whereas the <Heisenberg group> $B$ has $[I+E_{12},I+E_{23}]=I+E_{13}\ne I$.
To obtain arbitrarily many examples in dimension five, choose $r$ distinct odd primes $p_1,\ldots,p_r$ with $2^r\geq n$. For each prime use the preceding pair $E_j,B_j\leq S_{p_j^3}$. Set
$$
T=\prod_{j=1}^r S_{p_j^3},\qquad U_\varepsilon=\prod_{j=1}^r H_{j,\varepsilon_j},\quad
H_{j,0}=E_j,\ H_{j,1}=B_j,\quad\varepsilon\in\{0,1\}^r.
$$
A <conjugacy class> in a direct product is a product of <conjugacy classes>, and its intersection count with $U_\varepsilon$ is the product of the individual counts. Thus all $2^r$ subgroups are <Gassmann equivalent>. They are pairwise nonisomorphic: the unique Sylow $p_j$-subgroup of $U_\varepsilon$ is abelian precisely when $\varepsilon_j=0$, and an abstract <group isomorphism> preserves this property at each prime.
We use the <closed-manifold realization of a finitely presented fundamental group> in dimension five. Its construction here is short. From a finite presentation of $T$ with $s$ generators, start with the connected sum of $s$ copies of $S^1\times S^4$. Its <fundamental group> is free on those generators. Represent the finitely many relators by disjoint embedded circles, which is possible by <general position for curves in a manifold> in dimension five. Their oriented normal bundles are trivial. Perform framed circle surgery, removing $S^1\times D^4$ and inserting $D^2\times S^3$. The <Seifert-van Kampen theorem> shows that each surgery kills exactly the normal closure of the corresponding relator: the attaching circle bounds the new disk, and the new $S^3$ factor contributes no fundamental group. The resulting closed connected five-manifold $M_0$ has $\pi_1(M_0)=T$.
Choose any smooth <Riemannian metric> on $M_0$ and lift it to its <universal cover> $X$. Since $T$ is finite, $X$ is a compact simply connected five-manifold on which $T$ acts freely and isometrically. The <Sunada theorem> makes all $U_\varepsilon\backslash X$ isospectral, while
$$
\pi_1(U_\varepsilon\backslash X)\cong U_\varepsilon
$$
shows that they are pairwise nonhomeomorphic, hence nonisometric. Taking any $n$ members proves \b[for every positive integer $n$, there are $n$ pairwise nonisometric closed five-manifolds with the same Laplacian spectrum].
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