= Solution
Consider two finite assemblies of the same number $n$ of congruent bounded Euclidean tiles. Label their matching faces so the same prescribed face identifications are used in both assemblies. For each face type $c$, encode the first assembly by a symmetric involution $A_c$: a paired face exchanges its two tile indices, an exterior <Dirichlet boundary condition> contributes diagonal $-1$, and an exterior <Neumann boundary condition> contributes diagonal $+1$. Let $B_c$ encode the second assembly. Face coordinates are pulled back to the same reference face; if identifications have additional face isometries, these pullbacks must be included in the intertwining operators.
The <transplantation theorem> says that a constant matrix $C$ with $CA_c=B_cC$ for every face type sends the vector of tile restrictions of a <Laplacian eigenfunction> on the first assembly to one on the second, by $g=Cf$. If $C$ is invertible, it is a <bijection> of each <eigenspace> and the assemblies are isospectral with multiplicities.
To prove it, each component solves the same interior equation $\Delta f_j=\lambda f_j$, and a constant linear combination solves that equation too. On a face, let $u$ be the vector of boundary values and $v$ the vector of outward <normal derivatives>. All matching and exterior conditions are exactly
$$
A_cu=u,\qquad A_cv=-v.
$$
For an internal face these equations say that values agree and the two outward derivatives sum to zero. For diagonal $-1$ they impose zero value, and for diagonal $+1$ zero <normal derivative>. The intertwining identity gives $B_c(Cu)=Cu$ and $B_c(Cv)=-Cv$, so the transplanted functions match and satisfy the correct <boundary conditions>. Applying $C^{-1}$ proves the <eigenspace> <bijection>. At corners the same reasoning is interpreted in the finite-energy weak domain: no value jump and cancellation of normal flux prevent an extra distributional source. It gives the usual self-adjoint Dirichlet/mixed Laplacians on polygonal assemblies.
Here is an explicit <transplantation by reflection parity>. Take a rectangle of width $2a$ and height $b$, with the <Dirichlet boundary condition> on every exterior side. Cut it at its vertical midline into two congruent half-rectangles, pulling the right-hand half back by reflection. The midline gluing matrix is
$$
S=\begin{pmatrix}0&1\\1&0\end{pmatrix}.
$$
Replace the connected assembly by the disjoint union of two width-$a$, height-$b$ rectangles: the first has Dirichlet conditions on all sides; the second has Dirichlet on three sides and Neumann on the side representing the cut. Its corresponding matrix is $D=\operatorname{diag}(-1,1)$. The invertible, indeed orthogonal, matrix
$$
\boxed{C=\frac1{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix},\qquad CS=DC}
$$
intertwines the midline conditions. Every other face has matrix $-I$ on both assemblies, which automatically intertwines. The <transplantation theorem> proves equality of the complete <spectra>, although \b[one surface is connected and the other has two components with the required uniform and mixed conditions].
Equivalently, odd reflection <eigenfunctions> vanish on the cut, while even reflection <eigenfunctions> have zero <normal derivative> there. A direct separation-of-variables check gives the connected rectangle's <eigenvalues>
$$
\pi^2\left(\frac{m^2}{4a^2}+\frac{n^2}{b^2}\right),\qquad m,n\geq1.
$$
Even $m=2j$ gives the fully Dirichlet half-rectangle <spectrum>, and odd $m=2j+1$, $j\geq0$, gives the mixed half-rectangle <spectrum>. The split preserves every multiplicity, including coincidences between different pairs of indices.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-12-reflection-transplantation.png]
{title=Reflection splits a connected Dirichlet rectangle into a disjoint Dirichlet rectangle and a mixed Dirichlet-Neumann rectangle}
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