= Solution
Use the following <skew-symmetric matrix>, whose columns reduce modulo three to four of the specified nonzero <tetracode> words:
$$
S=\begin{pmatrix}0&1&1&1\\-1&0&-1&1\\-1&1&0&-1\\-1&-1&1&0\end{pmatrix},\qquad S^T=-S,\qquad S^2=-3I.
$$
Set $B_+=3I+S$, $B_-=-3I+S$, and define the full <Euclidean lattices> $L_\pm^0=B_\pm\mathbb Z^4$. Their determinants have absolute value $144$, since
$$
B_+^{-1}=\frac{3I-S}{12},\qquad B_-^{-1}=\frac{-3I-S}{12}.
$$
For arbitrary positive real numbers $a_1,\ldots,a_4$, put $D=\operatorname{diag}(\sqrt{a_1},\ldots,\sqrt{a_4})$ and $L_\pm=DL_\pm^0$. This is the <tetracode length-isospectral lattice construction>. We prove equality of the length multisets by an explicit <bijection>, not just equality of the shortest lengths.
Modulo three, both <lattices> have image $\operatorname{im}S=T$. The <matrix> $S$ has rank two over $\mathbb F_3$; its four column lines are precisely the four lines in $T$. Also $S^2=0$ modulo three, so $\ker S=\operatorname{im}S=T$. Each nonzero <tetracode> word has exactly one zero coordinate. Every $v\in L_\pm^0$ therefore has some coordinate divisible by three. Moreover, all four coordinates of $v$ have the same parity, because every entry of either $B_\pm$ is odd.
For $v\in L_+^0$, choose the least index $i$ with $3\mid v_i$ and set $w=v-2v_ie_i$. The <coordinate reflection bijection for tetracode lattices> will show that $w\in L_-^0$. First, $w\equiv v\pmod3$, so $Sw\equiv0\pmod3$. Second, positive-lattice membership implies $Sv\equiv3v\equiv-v\pmod4$. Hence
$$
(I-S)w\equiv2\bigl(v-v_ie_i+v_iSe_i\bigr)\pmod4.
$$
The expression in parentheses has $i$th coordinate zero and every other coordinate $v_j+v_iS_{ji}$ even, by the common parity and $S_{ji}=\pm1$. Thus $(I-S)w\equiv0\pmod4$. Together the congruences modulo three and four say $(-3I-S)w\equiv0\pmod{12}$, which is precisely $B_-^{-1}w\in\mathbb Z^4$.
Conversely, if $v\in L_-^0$, the same reflection has unchanged reduction modulo three, and now $Sv\equiv v\pmod4$. Thus
$$
(3I-S)(v-2v_ie_i)\equiv2\bigl(v+v_ie_i+v_iSe_i\bigr)\pmod4.
$$
Its $i$th coordinate is $4v_i$ and its other coordinates are even multiples of two, so it vanishes modulo four; it vanishes modulo three as well. The reflected vector is in $L_+^0$. The chosen index is unchanged by reflection, including for vectors reducing to zero, so this map is an <involution> between the two <lattices>. It preserves every $|v_j|$ and therefore preserves $\|Dv\|^2=\sum_j a_jv_j^2$. \b[The two lattices have the same length spectrum, including every multiplicity, for every positive choice of weights.]
To prove a genuine failure of <isometry>, choose
$$
(a_1,a_2,a_3,a_4)=(1,\sqrt2,\sqrt3,\sqrt5).
$$
These four numbers are linearly independent over $\mathbb Q$: changing the signs of the independent square roots in the <multiquadratic field> $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$ isolates each coefficient of a rational relation. If an <orthogonal transformation> $O$ maps $L_+$ to $L_-$, then
$$
P=D^{-1}OD=B_-UB_+^{-1}\in\operatorname{GL}_4(\mathbb Q),\qquad U\in\operatorname{GL}_4(\mathbb Z),
$$
and $P^T\operatorname{diag}(a_j)P=\operatorname{diag}(a_j)$. Comparing rational coefficients of the independent $a_j$ in the diagonal entries gives $P_{ji}^2=\delta_{ji}$. Consequently $P$ is diagonal with entries $\pm1$.
Such a coordinate sign change must preserve $T$ modulo three. For each zero coordinate $i$, there are exactly two opposite nonzero <tetracode> words. Preserving their line requires the other three signs to be equal. Taking two different zero coordinates forces all four signs to be equal. Thus $P=I$ or $P=-I$, either of which would imply $L_+^0=L_-^0$. But
$$
v=B_+e_1=(3,-1,-1,-1)^T,\qquad B_-^{-1}v=(-1/2,1/2,1/2,1/2)^T\notin\mathbb Z^4.
$$
This contradiction proves \b[$L_+$ and $L_-$ are not isometric]. Allowing affine <isometries> changes nothing: translating the image of zero back to zero reduces them to the orthogonal case.
For the <flat tori>, take period lattices equal to the <dual lattices>:
$$
X_\pm=\mathbb R^4/L_\pm^\vee.
$$
The standard <spectrum of a flat torus> lemma states that its <characters of a real torus> are indexed by the dual of its period lattice. Here they are $e^{2\pi i\langle v,x\rangle}$ for $v\in L_\pm$, with <Laplacian> <eigenvalue> $4\pi^2\|v\|^2$. These <characters> give a complete orthogonal basis, so our length-preserving <bijection> proves that $X_+$ and $X_-$ are <isospectral manifolds>. The second standard lemma is that an <isometry> of <flat tori> lifts to an affine <isometry> of their Euclidean universal covers. Its orthogonal part maps the period lattices to one another, and therefore maps their <dual lattices> to one another. This would give the impossible <isometry> $L_+\cong L_-$. Hence \b[$X_+$ and $X_-$ are isospectral non-isometric flat four-dimensional tori].
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