Consider the graph: $1$--$2$--$3$.
The graph Laplacian is $$L(\mathcal{G})=\begin{bmatrix}1& -1& 0\\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix}$$ Construct the polynomial in quadratic form: $$p(\hat{x})=\hat{x}^TL(\mathcal{G})\hat{x},$$ where $\hat{x} = [x_1\ \ x_2 \ \ x_3]^T$. Obviously, if we permute $x_1$ and $x_3$ (or equivalently, permute node $1$ and node $3$), we get $$p_{(13)}(\hat{x}) = p(\hat{x}).$$
Now replace $L(\mathcal{G})$ by $\exp(L(\mathcal{G}))$ and consider $$p(\hat{x})=\hat{x}^Te^{L(\mathcal{G})}\hat{x}.$$ I use MATLAB to test this case and I find that I still can permute $x_1$ and $x_3$ to get the same polynomial (symmetry).
Does this hold for any connected graph with graph automorphism? Is there any way to prove it?
Thanks!
Let $A$ be the matrix of a graph automorphism, which is a permutation matrix, hence orthogonal: $A^{-1}=A^T$. Since it is an automorphism, it preserves the Laplacian: $A^TLA = L$. Then, as you noticed, the quadratic form $p(x)=x^TLx$ is invariant under $A$:
$$p(Ax) = (Ax)^T L (Ax) = x^TA^TLAx = x^T L x$$
The same would be true for $q(x) = x^T \exp(L) x$, if $A^T \exp(L) A = \exp(L)$, by a similar argument.
By induction, $A^T L^k A = A^T L L^{k-1} A = A^T L (A A^T) L^{k-1} A = (A^T L A) (A^T L^{k-1} A) = L^k$.
(we inserted $AA^T$ since $AA^T=1$).
Now, compute $A^T \exp(L) A$:
$$A^T \exp(L) A = A^T \left(\sum \frac{L^k}{k!}\right)A = \sum \frac{A^TL^kA}{k!} = \sum \frac{L^k}{k!} = \exp(L)$$
and the result follows.