Imagine you have a rigid rod which is free to move and a force is applied at a point away from the center of mass of the rod. This would create torques at multiple points on the rod as the force will have a line of action. Will this mean the rod will rotate around all these points due to the same torque? Won't this create multiple axis and violate the rigid property of rod?
How do we decide around which axis will the rotation start?
The rod accelerates as if all the forces acting on it are acting upon the center of mass. Thus, if the rod is free to move on a frictionless surface, the rod will still obey the equation $$\vec F = m\vec a$$ Again, if the rod is free to move (not constrained to rotate about a certain axis, and no other force acting on it), it will always rotate about its center of mass. That is, if the force is applied at a distance $r$ from the center, we have $$Fr = I\alpha$$ where $I = \frac{1}{12}mL^2$ is the moment of inertia about the center of mass, and $\alpha$ is the angular acceleration.
The translational motion and the rotational motion can be separated. Imagine the rod as a point mass (that is, a dot) accelerating with $\vec F = m\vec a$, and rotating about it.
The laws of motion state the following:
An eccentric load-line causes the center of mass the move linearly (in line with the force) and the body to rotate about the center of mass due to the moment arm of the force w.r.t. the center of mass.
The combined motion yields an instant center of rotation that is "on the other side" of the center of mass as the applied load and at a distance from the center of mass. The distance of the instant center of rotation to the center of mass is $$ c = \frac{I}{m\, d}$$
Here $m$ is the mass, $I$ is the mass moment of inertia and $d$ the moment arm of the load.
Specifically, with your question, each (small) part of the body has only linear motion and no rotation (a point cannot rotate). The rotation of a rigid body is an emergent property from the combined motion of all the particles in the body. This becomes clear when you can derive the equation of motion of a rigid body from the definition of momentum of each individual particle.
Imagine a mass described by a density function $\rho({\vec r})$ rotating about a point ${\vec r}_0$ with angular velocity $\vec{\omega}$. The linear momentum of the mass is:
$${\vec p}=\int{\rho({\vec r})\vec v(\vec r)d^3r} $$
where the velocity of a point is given by:
$$\vec v(\vec r)=\vec\omega\times[\vec r-\vec r_0] $$
So:
$$\vec p =\int{\vec{\omega}\times[\vec r - \vec r_0]\rho({\vec r})d^3r} $$
Since there are no external forces:
$$ \vec p = 0$$
Hence:
$$ \int{[\vec\omega\times\vec r]\rho(\vec r)d^3r}=\int{[\vec\omega\times\vec r_0]\rho(\vec r)d^3r} $$
At this point, you can solve for $\vec r_0$, or plug in the center-of-mass:
$$ \vec r_{com}\equiv \frac{\int{\vec r\rho(\vec r)d^3r}}{\int{\rho(\vec r)d^3r}}$$
and verify that it is the solution.
