Does the Sun orbit a much larger nearby star? EDIT: I got thumbed down, so I removed the details. The question is already crystal clear, IMO.
 A: To find out if two objects form a gravitationally bound system we compare their kinetic and potential energies. If the kinetic energy is less than the potential energy the system is bound, while if the kinetic energy is higher than the potential energy the system is not bound.
The nearest big star is Alpha Centauri at 4.35 light years, so let's do the calculation for the Sun and Alpha Centauri. The potential energy is:
$$ U = \frac{G M_1 M_2}{d} $$
where $d$ is the distance between the stars and $M_1$ and $M_2$ are the masses of the Sun and Alpha Centauri respectively. Alpha Centauri is about 1.1 solar masses. For a rough calculation (we should really work in the centre of mass frame) let's take the Sun as fixed, and work out what velocity for Alpha Centauri gives a matching kinetic energy:
$$ \frac{1}{2} M_2 v^2 = \frac{G M_1 M_2}{d} $$
so:
$$ v = \sqrt{\frac{2 G M_1}{d}} $$
Feed in the mass of the Sun and the distance and we get:
$$ v \approx 80 ms^{-1} $$
So the Sun could only be bound in orbit round Alpha Centauri if it's relative velocity is less than about 80 m/s. However the radial relative velocity is 21,600 m/s. So the Sun and Alpha Centauri are not in orbit round each other. You could, with enough patience, repeat this calculation for all nearby stars and convince yourself the Sun isn't gravitationally bound to any of them.
The Sun is of course in orbit round the combined mass of all the stars in the Milky Way, but then the total mass of the Milky Way is about $10^{12}$ solar masses. This means a star is bound to the Milky Way if its velocity is less than about 700-800 m/s, and the Sun's velocity relative to the Milky Way is about 210 m/s.
