jmc

We know that the graphs of $y=f(x)$ and $y=f^{-1}(x)$ are reflections of each other across the line $y=x.$ If they intersect at some point $(a,b),$ where $a\ne b,$ then they must also intersect at the point $(b,a),$ which is the reflection of the point $(a,b)$ in the line $y=x.$

But we are told that the graphs have exactly one point of intersection, so it must be of the form $(a,a).$ Since this point lies on the graph of $y=f(x),$ $a=f(a).$ In other words, $$a=a^3+6a^2+16a+28.$$Then $a^3+6a^2+15a+28=0,$ which factors as $(a+4)(a^2+2a+7)=0.$ The quadratic factor does not have any real roots, so $a=-4.$ The point of intersection is then $\boxed{(-4,-4)}.$