The theta function of a lattice $L$ is the generating function $\theta_{L}(q)$ counting the number of lattice points of a given squared-length in $L$: $$\theta_{L}(q):=\sum_{x\in L}q^{x\cdot x}=\sum_{\alpha\in\left\{ x\cdot x\mid x\in L\right\} }N_{\alpha}q^{\alpha},\quad N_{\alpha}:=\#\left\{ x\in L\bigm|x\cdot x=\alpha\right\} .$$$$\theta_{L}(q):=\sum_{x\in L}q^{x\cdot x}=\sum_{\alpha\in\left\{ x\cdot x\mid x\in L\right\} }N_{\alpha}q^{\alpha},\quad N_{\alpha}:=\#\left\{ x\in L\bigm|x\cdot x=\alpha\right\}.$$ (Commonly there is a factor of $\tfrac{1}{2}$ in the exponent.)
$$ D_{n}^{+} :=D_{n}\cup\left(D_{n}+\tfrac{1}{2}c_{n}\right). $$$$ D_{n}^{+}:=D_{n}\cup\left(D_{n}+\tfrac{1}{2}c_{n}\right). $$
Both $\mathbb{Z}^{n}$ and $D_{n}^{+}$ have$D_{n}$ has index $2$ in both $D_{n}$$\mathbb{Z}^{n}$ and $D_{n}^{+}$. ThusSince $\det\mathbb{Z}^{n}=1\implies\det D_{n}=2^{2}\implies\det D_{n}^{+}=1$$\det\mathbb{Z}^{n}=1$ and $\det(L')/\det(L)=[L:L']^2$, soit follows that $D_{n}^{+}$ is unimodular$\det D_{n}=4$ and $\det D_{n}^{+}=1$.
In Four-Dimensional Lattices With the Same Theta Series, Conway and Sloane construct an explicit length-preserving bijection between isospectral lattices (changing the sign of the first coordinate which is divisible by 3). I'm curious if a similar construction is possible in the above case.
EDIT: Attempted solution
Perhaps there is some clever way to slice up, reflect/rotate, and recombine the two lattices into each other.
One possible way to slice them up is into cosets. For this we need a common sublattice. An obvious candidate is the index 4 sublattice $D_8\oplus D_8$ given by
$$ \begin{align*} 0\rightarrow D_{8}\oplus D_{8} & \rightarrow D_{8}^{+}\oplus D_{8}^{+}\rightarrow\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\rightarrow0,\\ 0\rightarrow D_{8}\oplus D_{8} & \rightarrow D_{16}^{+}\rightarrow\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\rightarrow0. \end{align*} $$
The rightmost map on the top is obvious, while the rightmost map on the bottom measures
- the parity of the sum of the first eight coordinates
- whether $\tfrac{1}{2}$ appears in some/every coordinate.
Then we can start computing the theta functions of the cosets. Recall the standard formulas $\theta_{D_n}=\tfrac{1}{2}(\theta_3^n+\theta_4^n)$ and $\theta_{D_n^+}=\tfrac{1}{2}(\theta_2^n+\theta_3^n+\theta_4^n)$ (see SPLAG 2.3.37).
Denoting the theta functions of the $\mathbb{Z}_2\oplus\mathbb{Z}_2$ cosets $(a,b)$ with superscripts $\theta^{a,b}$,
\begin{align*} \theta_{D_{8}^{+}\oplus D_{8}^{+}}^{0,0}=\theta_{D_{16}^{+}}^{0,0}=\theta_{D_{8}}^{2} & =1+224q^{2}+14816q^{4}+260736q^{6}+\cdots,\\ \theta_{D_{8}^{+}\oplus D_{8}^{+}}^{0,1}=\theta_{D_{8}^{+}\oplus D_{8}^{+}}^{1,0}=\theta_{D_{8}}(\theta_{D_{8}^{+}}-\theta_{D_{8}}) & =128q^{2}+15360q^{4}+263680q^{6}+\cdots,\\ \theta_{D_{8}^{+}\oplus D_{8}^{+}}^{1,1}=(\theta_{D_{8}^{+}}-\theta_{D_{8}})^{2} & =16384q^{4}+262144q^{6}+\cdots,\\ \theta_{D_{16}^{+}}^{0,1}\stackrel{?}{=}\theta_{D_{16}^{+}}^{1,1}\stackrel{?}{=}(\theta_{D_{8}^{+}}-\theta_{D_{8}})^{2} & =16384q^{4}+262144q^{6}+\cdots,\\ \theta_{D_{16}^{+}}^{1,0}\stackrel{?}{=}(\theta_{D_{8}^{+}}-\theta_{D_{8}})(3\theta_{D_{8}}-\theta_{D_{8}^{+}}) & =256q^{2}+14336q^{4}+265216q^{6}+\cdots. \end{align*}
The last two lines were done with a computer up to $q^8$, hence marked with "$\stackrel{?}{=}$".
The theta functions of the cosets don't match up, but they are related by very simple linear dependencies, so it seems plausible that this strategy may eventually succeed. (For example, note that $3\theta_{D_{8}}-\theta_{D_{8}^{+}}=\theta_{D_{4}^2}$.)