If an object starts spinning as a result of some internal torque, would its linear velocity change? If an object is moving at a constant velocity, and starts spinning as a result of some internal torque, would its linear velocity decrease or stay the same?
Since no external forces are applied, the objects total energy should stay the same. This means the sum of its translational kinetic, rotational kinetic, and internal energies should not change. Since rotational kinetic energy increases, one of the other types must decrease. I am thinking that since the rotation requires a torque to get started, this means some form of internal energy must be used, so internal energy would decrease, and translational kinetic energy would stay the same. Is this correct?
 A: I am trying to understand your question, for example, consider a satellite which has a couple of lateral burners. When they are started together the satellite rotates and that is all. If only one or them is started (as in the first part of the figure) you have both rotation and translation, changing internal, rotational and translational kinetic energies at the same time in the process. A torque couple is needed to produce just rotation so the net force is cancelled but not the torque as in the second case or the figure. If the force due to some internal process then a change in the internal energy is needed.
 
A: Suppose a ship drifting in outer space that is no more than a running wheel with a hamster inside. The hamster is initially at rest.
Suddenlly it starts to run. As angular momentum is conserved, the hamster and the running wheel will rotate in opposite directions. The linear momentum doesn't change. Mechanical energy increases, but at expenses of the internal energy of the hamster.
If there is some jet device to turn the ship, both linear and angular momentum can change, but only because we are forgetting the exausting gases. Keeping the hypothesis of an internal process, they should be include in the calculation. And everything is conserved.
A: Let's take a system of N discrete particles. It wouldn't matter if they were continuously connected which you will see.  The total angular momentum is: 
$$\overrightarrow L=\sum_{i=1}^N \overrightarrow r_i×\overrightarrow p_i$$
Now differentiate both sides with respect to time. 
$$\frac{d\overrightarrow L}{dt}=\frac{d}{dt}\sum_{i=1}^N \overrightarrow r_i×\overrightarrow p_i$$
$$=\sum_{i}{\frac{d\overrightarrow r_i}{dt}}×\overrightarrow p_i +\sum_{i}{\overrightarrow r_i}×\frac{d\overrightarrow p_i}{dt}$$
$$=\sum_{i}{\overrightarrow v_i}×(m\overrightarrow v_i)+\sum_i \overrightarrow {r_i} ×(\overrightarrow F_i^{ext} +\overrightarrow F_i^{int})$$
$$=0+\sum_i \overrightarrow r_i ×\overrightarrow F_i^{ext}$$
$$≡\sum_i \overrightarrow \tau _i^{ext}$$
Second to last line follows because $\overrightarrow v_i × \overrightarrow v_i =0$ and also $\sum_i \overrightarrow r_i × \overrightarrow F_i^{int}=0$. In other words, the internal forces exerted by particles on each other provide no net torque. This makes sense because a rigid object with no external forces won't spontaneously start rotating. 
As everyone is using the example of space ships I would like to point out that the net torque is still 0 for fuel+rocket system about the same axis. The fuel ejected will have the opposite torque as the rocket. One can also consider the rocket minus fuel as their system which will make the force applied external and net internal torque is still 0. Obviously energy for rotation comes from the stored potential energy of the fuel. 
