Under what conditions does matrix multiplication commute? This is just a check on my reasoning, I guess.
So for two matrices $A, B$ to commute, the following must hold:
$$(AB)_{ij} = \sum_{k=1}^{n}a_{ik}b_{kj} = \sum_{k=1}^{n}b_{ik}a_{kj} = (BA)_{ij}$$
This can happen if for all $i, j, k$:
a. $a_{ik}=a_{kj}$ and $b_{ik}=b_{kj}$, or
b. $a_{ik}=b_{ik}$ and $a_{kj}=b_{kj}$, i.e. $A=B$, or
c. $a_{ik} = 0$ or $b_{ik} = 0$, i.e. either matrix is null.
Are there more possibilities? 
Edit: I originally had (a) as "Both matrices are symmetric", but as @user1551 points out, this is not true. After fixing the summations, I see where I was mistaken. I'm not sure how to characterize (a) now.

Two matrices commute when they are simultaneously triangularisable, i.e., when there is some basis in which they are both triangular. Roughly speaking, it is when they have the same eigenvectors, probably with different eigenvalues. (But then there are degenerate cases, which make it all more complicated.)
This property has really nothing to do with A and B being symmetric. Indeed, there are examples of matrices which are symmetric and don't commute...
$$A=\left(\begin{matrix}2& 1\\1 & 3 \end{matrix}\right),\;
B=\left(\begin{matrix}3& 1\\1 & 2 \end{matrix}\right),
$$
...and those which are not symmetric but do commute:
$$A=\left(\begin{matrix}1& 1\\0 & 1 \end{matrix}\right),\;
B=\left(\begin{matrix}1& 2\\0 & 1 \end{matrix}\right).
$$
