Paragraph copied from Bona.
"Let $f \in Aut(H)$ and let A be a vertex of the cube. Then we have 8 choices for $f(A)$. Once $f(A)$ is chosen, the neighbors of $A$ in $H$ must be mapped into the neighbors of $f(A)$ by $f$. Since $A$ has three neighbors, this is possible in $3!$ ways. Once the images of the neighbors are chosen, there are no more choices to make (why?)"
I played around by drawing pictures and it is clear that the other vertices are fixed once $f(A)$ and neighbors of $f(A)$ are chosen. But why? How does one prove that the others will be fixed and will not contribute to the total automorphisms?
Thanks