What's the derivative of degree (Laplacian) matrix with respect to adjacency matrix?
For a symmetric adjacency matrix $A\in \{0, 1\}^{n \times n}$, the degree is defined as $D_{ii} = \sum_i A_{ij} = \sum_j A_{ij}$.
What the derivative of the degree matrix w.r.t. adjacency matrix?
I know it's a 4-dimensional tensor. Taking $n=3$ as example
$$ \frac{\partial D_{00}}{\partial A} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ or
$$ \frac{\partial D_{00}}{\partial A} = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} $$
or $$ \frac{\partial D_{00}}{\partial A} = \begin{pmatrix} 1 & 1/2 & 1/2 \\ 1/2 & 0 & 0 \\ 1/2 & 0 & 0 \end{pmatrix} $$ or $$ \frac{\partial D_{00}}{\partial A} = \begin{pmatrix} 0 & 1/2 & 1/2 \\ 1/2 & 0 & 0 \\ 1/2 & 0 & 0 \end{pmatrix} $$
which one is correct?
Ideally, it should be symmetric, and the element on the diagonal should be 1 or 0?
The derivative you are looking has definitely to be symmetric.
Let $\phi=D_{00}=\mathrm{tr}(A_{00}+A_{01}+A_{02})$ and consider the perturbation $$ d\mathbf{A}= \begin{pmatrix} 0 & 0 & h \\ 0 & 0 & 0 \\ h & 0 & 0 \end{pmatrix} $$ We know that this pertubation should yield to $d\phi=h$.
The relation $d\phi= \frac{\partial D_{00}}{\mathbf{A}}:d\mathbf{A}$ is correct iff $$ \frac{\partial D_{00}}{\mathbf{A}} = \begin{pmatrix} 1 & 1/2 & 1/2 \\ 1/2 & 0 & 0 \\ 1/2 & 0 & 0 \end{pmatrix} $$
$ \def\c#1{\color{red}{#1}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\o{{\large\tt1}}\def\d{{\theta}}\def\e{{\large\varepsilon}} \def\E{{\cal E}}\def\F{{\cal F}}\def\G{{\cal H}} \def\L{\left}\def\R{\right} \def\LR#1{\L(#1\R)} \def\BR#1{\Big(#1\Big)} \def\bR#1{\big(#1\big)} \def\vecc#1{\operatorname{vec}\LR{#1}} \def\Sym#1{\operatorname{Sym}\LR{#1}} \def\diag#1{\operatorname{diag}\LR{#1}} \def\Diag#1{\operatorname{Diag}\LR{#1}} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\p{\partial}\def\grad#1#2{\frac{\p #1}{\p #2}} $Use the all-ones vector $\{\o\}$ and the tensors $\{\d,\G\}$ with components $$\eqalign{ \d_{ijk} &= \begin{cases} \o\quad{\rm if}\;(i=j=k) \\ 0 \quad{\rm otherwise} \\ \end{cases} \\ \G_{ijk\ell} &= \begin{cases} \o\quad{\rm if}\;(i=j)\;{\rm and}\;(k=\ell) \\ 0 \quad{\rm otherwise} \\ \end{cases} \\ }$$
to write the degree matrix in the following form and calculate its gradient $$\eqalign{ D &= \Diag{A\cdot\o} &= \d\cdot\bR{A\cdot\o} \\ dD &= \bR{\d\cdot dA\cdot\o} &= \bR{\d\cdot\G\cdot\o}:dA \\ \grad{D}{A} &= \bR{\d\cdot\G\cdot\o} &= \bR{\d\star\o} \qquad \big({\rm "\!\!the\;gradient\!\!"}\big) \\ }$$ where $(\star)$ denotes dyadic product, $(\cdot)$ the single dot product, and $(:)$ the double dot product.
The gradient is a $4^{th}$order tensor, which admits 2 different decompositions into matrix-valued components. Multiplying on the left by the Cartesian basis matrix $E_{ij}$ produces the components $$\eqalign{ &E_{ij}:\LR{\grad{D}{A}} = E_{ij}:\bR{\d\star\o} \\ &\c{\grad{D_{ij}}{A}} = \diag{E_{ij}}\cdot\o^T \\ }$$
while multiplying on the right produces components $$\eqalign{ &\LR{\grad{D}{A}}:E_{ij} = \bR{\d\star\o}:E_{ij} \\ &\grad{D}{A_{ij}} = \bR{\d\cdot E_{ij}\cdot\o} = \bR{\d\cdot\e_i} = \Diag{\e_i} = E_{ii} \\ }$$ where $\e_i$ is a Cartesian basis vector and $\,E_{ij}=\e_i\e_j^T$
So in your example, the first matrix is the correct one since
$$
\c{\grad{D_{00}}{A}}
= \diag{E_{00}}\cdot\o^T
= \m{1\\0\\0}\cdot\m{1&1&1}
= \m{1&1&1\\0&0&0\\0&0&0}
$$
Note that Diag(v) converts a vector into a diagonal matrix,
while diag(M) extracts the diagonal of a square matrix into a vector.
Oops! I forgot the symmetry constraint on $A$. But that merely symmetrizes the gradient, so $$ \Sym{\grad{D_{00}}{A}} = \frac 12\LR{\m{1&1&1\\0&0&0\\0&0&0}+\m{1&1&1\\0&0&0\\0&0&0}^T} = \frac 12 \m{2&1&1\\1&0&0\\1&0&0} $$ which is your third matrix.