= Solution
Take a finite presentation $T=\langle x_1,\ldots,x_r\mid R_1,\ldots,R_s\rangle$. For a <closed-manifold realization of a finitely presented fundamental group>, begin with a five-dimensional ball and attach $r$ one-handles, giving a compact oriented five-manifold $W_1$ with <fundamental group> the free group on the $x_i$. Represent the relators by disjoint embedded circles in its four-dimensional boundary. General position allows them to be embedded and disjoint, and their oriented normal bundles admit framings. Attach a two-handle along each such circle. The resulting compact five-manifold $W$ has
$$
\pi_1(W)=\langle x_1,\ldots,x_r\mid R_1,\ldots,R_s\rangle=T
$$
by the <Seifert-van Kampen theorem>. Now set $B=\partial W$. It is a compact connected smooth four-manifold without boundary. Turning this <handle decomposition> upside down gives a relative handle decomposition of $(W,B)$ with handles of indices three, four and five only. Attaching such handles does not change the <fundamental group>: there are no new one-handles or two-handle relations. Thus $\pi_1(B)\to\pi_1(W)$ is an isomorphism and
$$
\boxed{\pi_1(B)\cong T.}
$$
These are the differential-topological handle facts used in the construction; the dimension-five thickening ensures the boundary retains the prescribed <fundamental group>.
Now assume $T$ finite. Choose any smooth <Riemannian metric> on $B$ and lift it to its <universal cover> $\widetilde B$. This cover is compact, since its degree is $|T|$, and simply connected. Its <deck transformations> give a free isometric $T$-action. For the two <Gassmann equivalent> subgroups put
$$
M_i=U_i\backslash\widetilde B.
$$
Each is a closed <Riemannian manifold>, and $\widetilde B$ is its simply connected universal cover, with <deck transformation group> $U_i$. Consequently $\pi_1(M_i)\cong U_i$.
Here is the complete spectral argument behind the <Sunada theorem> in this situation. Each finite-dimensional upstairs <eigenspace> $E_\lambda$ is a representation of $T$, because the action is isometric and commutes with the <Laplace-Beltrami operator>. Functions downstairs correspond exactly to invariant functions upstairs, so their multiplicities are $\dim E_\lambda^{U_i}$. Averaging the representation matrices over a subgroup is a projection onto its invariant subspace. Taking its <trace> gives
$$
\dim E_\lambda^{U_i}=\frac1{|U_i|}\sum_{u\in U_i}\chi_\lambda(u).
$$
The <character> $\chi_\lambda$ is constant on each <conjugacy class> of $T$. <Gassmann equivalence> means $|U_1\cap C|=|U_2\cap C|$ for every such class $C$; summing these equalities also gives $|U_1|=|U_2|$. The two averages are therefore equal for every $\lambda$. \b[The quotients are isospectral, with $\pi_1(M_i)\cong U_i$.]
If $U_2=tU_1t^{-1}$, the map $[x]_{U_1}\mapsto[tx]_{U_2}$ is well defined, invertible and isometric because $t$ acts by an <isometry> upstairs. \b[Conjugate subgroups give isometric quotients.]
For an explicit example with different <fundamental groups>, use the <order-sixteen Gassmann pair>
$$
H_1=C_4\times C_4,\qquad H_2=Q_8\times C_2.
$$
Embed each in $T=S_{16}$ by left multiplication on its own sixteen elements, identified with the same sixteen-point set. These <regular representations> are faithful. Both groups have one identity element, three elements of order two, and twelve of order four: in $C_4^2$ the elements of order dividing two are precisely its four even-coordinate elements; in $Q_8\times C_2$ they are the four elements in $\{1,-1\}\times C_2$. Every remaining element has order four.
In a regular action, an order-$k$ element is a product of $16/k$ disjoint $k$-cycles. Thus each subgroup intersects the identity <conjugacy class> of $S_{16}$ in one element, the class of type $2^8$ in three, the class of type $4^4$ in twelve, and every other class in zero. They are <Gassmann equivalent>. However $H_1$ is abelian and $H_2$ is nonabelian, so they are not isomorphic. Apply the preceding construction with <fundamental group> $S_{16}$ and these subgroups. It produces
$$
\boxed{\operatorname{Spec}(M_1)=\operatorname{Spec}(M_2),\qquad
\pi_1(M_1)\cong C_4^2\not\cong Q_8\times C_2\cong\pi_1(M_2).}
$$
In particular they cannot even be homeomorphic, and therefore cannot be isometric.
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