Do objects rotate around the torque vector or its center? If a sphere has a torque vector coming out of it at point A, would the sphere rotate about its center or the axis of the torque vector?
 A: The motion will depend on the forces that the torque arises from in the first place.
Newton's second law must apply to the system's centre of mass and is of course reckoned with the nett force. 
So if this nett force is nought, the centre of mass either cannot shift or moves uniformly (if so we shall assume the sphere's centre of mass is stationary relative to us). Therefore the centre of mass - the sphere's centre - must be on the axis of rotation. The rotation axis is therefore through the sphere's centre. 
Furthermore, in this zero nett force case, a torque vector's tail's position has no physical meaning because it can be shown then that the computed torque owing to a system of forces is independent of where one does the calculation (i.e. the point about which one calculates moments of forces), even though, of course, the torques owing to each individual force depend a great deal on where the calculation is done.
The notion that one can apply Newton's second law to a rigid body and calculate the path of the body's centre of mass as though the body were a point is known sometimes as Euler's First Law.
Euler's Second Law is that the torque $\mathbf{M}$ and the rigid body's angular momentum $\mathbf{L}$ are related analogously with Newton's second law, to wit:
$$\mathbf{M} = \mathrm{d}_t \mathbf{L}$$
and $\mathbf{L} = \mathbf{I}\, \vec{\omega}$ where $\mathbf{I}$ is now the moment of inertia tensor (a symmetric $3\times3$ matrix) so that $\mathbf{L}$ is not always in the same direction as the angular velocity $\vec{\omega}$. If there is a nett force, then the nett torque $\mathbf{M}$ depends on where it is calculated, but then so do $\mathbf{I}$ and $\mathbf{L}$, so that the motion's description is of course independent of the reference point, as it must be for physics cannot depend on the standpoint which humans describe it from! Moreover, $\mathbf{I}$ changes as the body is rotated by nett torques, so that it is more useful to do rigid body rotation dynamics calculations in a non-inertial frame that rotates with the body. If the frame is aligned along the princpal axes of inertia, i.e. the orthogonal eigenvectors of the inertia matrix, then the inertia matrix is diagonalised and Euler's second law takes on its simplest expression through the Euler equations. All analysis is wontedly done with the origin at the body's centre of mass, whose path is particularly simple owing to Euler's first law. 
For a sphere, the inertia matrix when calculated through any orthogonal axes through its centre is diagonal and "scalar" (i.e. proportional to the identity matrix), so we have the simple relationship $\mathbf{M} = I \mathrm{d}_t \vec{\omega}$. Angular momentum and angular velocity are always in the same direction. Torque begets rotation about an instantaneous axis through the centre of mass, and this rotation is superimposed on accelerated translation if the forces are unbalanced.
A: If a body moves only because of the influence of a torque, then it will rotate about the center of mass.
There is no location for torques, only directions. You you take the equations of motion as seen here (https://physics.stackexchange.com/a/80449/392) you will see that the location of the torque does not enter into the equations. Only the location of the forces.
As a result the acceleration of the center of mass is zero, and only angular velocity will exist. The body will rotate about its center of mass. 

Note that these two statements are equivalent:
  
  
*
  
*A pure force thorugh the center of mass (with no net torque about the center of mass) will purely translate a rigid body (any point on the body).
  
*A pure torque any point on the body (with no net force) will purely rotate a rigid body about its center of mass.
  

Consider a motionless rigid body with a pure instantenous torque $\vec{\tau}$ applied on it. The motion of any point A not on the center of mass is
$$ 0 = m \vec{a}_A - m \vec{c}\times \vec{\alpha} 
\\ \vec{\tau} = I_c \vec{\alpha} - m \vec{c} \times \vec{c} \times \vec{\alpha} + m \vec{c} \times \vec{a}_A $$
where $\vec{c}$ the position vector of the center of gravity relative to point A. The soltution to the above is 
$$ \vec{a}_A = \vec{c} \times \vec{\alpha} 
\\ \vec{\tau} = I_c \vec{\alpha}- m \vec{c} \times \vec{c} \times \vec{\alpha}+ m \vec{c} 
\times \vec{c} \times \vec{\alpha} = I_c \vec{\alpha} $$
$$ \vec{\alpha} = I_c^{-1} \vec{\tau} \\  \vec{a}_A = \vec{c} \times I_c^{-1} \vec{\tau} $$
From the above is it obvious that the only point A not moving is at $\vec{c}=0$ and all points parallel to $ \vec{\alpha}$ through the center of mass.
