Why does an unhinged body rotate about its centre of mass?

It happens I think because rotation about axes which do not pass through the center of mass are usually inherently unstable and hence the rotation tends to "decay" to a more stable axis, i.e the one passing through its center of mass. Also, the moment of inertia of a body is usually the lowest through a certain principle axis through its center of mass.(Moment of inertia is analogous to mass in rotation. i.e. more moment of inertia for a certain amount of force force implies lesser angular acceleration) A simple proof for this is:

The moment of inertia $I=mr^2$ for a point particle at a distance r from its axis of rotation. Now consider two particles separated by some distance $l$. Assume an axis of rotation passing through the line joining these both, at a distance $x$ from the first particle, and $l-x$ from the second. then the moment of inertia about this axis is $$I=Mx^2 + M(l-x)^2$$ Differentiating with respect to $x$, we get: $$\frac{dI}{dx}=2Mx-2Ml+2Mx$$ If you want to find the minimum value of x needed for the lowest of moment of inertia, equate the expression above to zero. This gives $I$ to be minimum at $$x=l/2$$ from the first particle which is the position of center of mass of the system. The moment of inertia is always lowest through an axis through the center of mass. (True for all rigid bodies.) and hence the most stable rotation is achieved through an axis through the center of mass.

The idea is that if there are no forces on an object, then no matter how it rotates, its center of mass must move at a constant velocity. Then in the frame of the object, the center of mass appears stationary and everything else rotates around it. In general this cannot be said of any other point in the object.

To see that the center of mass moves at a constant velocity, remember that the center of mass $\renewcommand{\r}{\mathbf{r}} \renewcommand{\rcm}{\r_\textrm{cm}} \rcm$ is defined by $\rcm = \frac{1}{M}\int \r \rho d \r$, where $\rho$ is the mass density distribution of the object and $M = \int \rho d \mathbf{r}$ is the total mass of the object. Then the center of mass velocity $\renewcommand{\a}{\mathbf{a}} \renewcommand{\acm}{\a_\textrm{cm}} \acm$ can be found by taking two time derivatives: $\acm = \frac{1}{M}\int \a \rho d \r$.

Now the integrand $\a(\r) \rho(\r)$, by newton's second law must be the total force $\renewcommand{\F}{\mathbf{F}}\F$ at the point $\r$. This force has two contributions: an external force $\renewcommand{\Fext}{\F_\mathrm{ext}}\Fext$ and an internal force $\renewcommand{\Fint}{\F_\mathrm{int}}\Fint$. Thus we have that $\a(\r) \rho(\r) = \F(\r) = \Fint(\r) + \Fext(\r)$. Plugging this back into our expression for $\acm$, we find $\acm = \frac{1}{M} \int \Fint(\r)+\Fext(\r) d \r = \frac{1}{M} \Fext + \frac{1}{M} \int \Fint(\r) d\r$, where $\Fext$ is the total external force.

Now the internal force comes from pairwise interactions with other parts of the object. So if $\Fint(\r, \r')$ is the force of the piece of the object at $\r'$ on the piece of the object at $\r$, then the total internal force at $\r$ is given by $\Fint(\r)=\int \Fint(\r,\r') d \r'$. Then the total contribution of $\Fint$ to $\acm$ can be written $\frac{1}{M} \int \Fint(\r) d\r = \frac{1}{M} \int \Fint(\r,\r') d\r' d\r$. But switching the order of integration, we find $\frac{1}{M} \int \Fint(\r,\r') d\r' d\r = \frac{1}{M} \int \Fint(\r,\r') d\r d\r'$ by Newton's third law we have $\Fint(\r,\r')=-\Fint(\r',\r)$. Combining this with the previous equation we find $\frac{1}{M} \int \Fint(\r,\r') d\r' d\r = -\frac{1}{M} \int \Fint(\r',\r) d\r d\r'$. But then relabelling the dummy variables on the right hand side, we find that $\frac{1}{M} \int \Fint(\r,\r') d\r' d\r = -\frac{1}{M} \int \Fint(\r,\r') d\r' d\r$. which implies that $\frac{1}{M} \int \Fint(\r,\r') d\r' d\r =0$. Therefore $\frac{1}{M} \int \Fint(\r) d\r =0$, and so $\acm = \frac{1}{M} \Fext $. Therefore if the external force is zero, the center of mass moves with constant velocity.