It's an exercise to check it coincides with $\mathrm{Aut}(S_6)$ which has $S_6$ as subgroup of index 2.

Here are the steps. First, for arbitrary $n\ge 6$, consider the graph of transpositions. So this is the set $X_n$ of the $n(n-1)/2$ unordered pairs, with an edge between two whenever they are disjoint.

On $X_n$, consider the set $Y_n$ all unordered pairs of non-joined vertices (hence of the form $\{\{a,b\},\{a,c\}$ for $a,b,c$ pairwise distinct. There's a canonical map $\phi$ $Y_n\to\{1,\dots,n\}$ given by $\phi(\{\{a,b\},\{a,c\})=a$; we wish to show it's equivariant for the group action.

Link two elements of $Y_n$ if they are disjoint and contained no pair of joined vertices. Then any two linked elements of $Y_n$ have the same image by $\phi$ (check! this uses the disjointness assumption). Next consider the equivalence relation on $Y_n$ generated by this equivalence relation. The only 2-element subsets in $Y_n$ with same image by $\phi$ but not linked have the form, up to permutation, $\{\{12,13\},\{12,14\}\}$. But one indeed have $\{12,13\}-\{15,16\}-\{12,14\}$. So for $n\ge 6$, the equivalence relation on $Y_n$ generated by being linked consists of being in the same fibers of $\phi$, and hence the automorphism group of $X_n$ acts on $\{1,\dots,n\}$. This action is easily seen to be faithful. Since the permutation group $S_n$ already acts, this shows that the automorphism group of $X_n$ is $S_n$.

Next in $S_n$ for $n\ge 7$, the transpositions are the only elements whose centralizer has order $2(n-2)!$, i.e., form the only conjugacy class of order $n(n-1)/2$, and hence $\mathrm{Aut}(\mathrm{Comm}(S_n))$ stabilizes its subgraph $X_n$. Again a little argument shows that this action is faithful (i.e. an automorphism of $\mathrm{Comm}(S_n)$ fixing pointwise the transpositions is the identity; also for $n=6$.

For $n=6$ on the other hand, there are 2 conjugacy classes of order 15: transpositions and triple transpositions. They are switched by non-inner automorphisms. Hence the previous argument applies to the subgroup of index 2 of $\mathrm{Aut}(\mathrm{Comm}(S_6))$ stabilizing $S_6$ and the result follows.