if mass is assumed to be constant, the velocity of the centre of mass of the system has to be different after the collision for the kinetic energy to be different.
However, if the momentum of the system is conserved, the velocity of the centre of mass of the system should remain the same.
- mass is not constant and velocity is different: in a completely inelastic collision the two objects (A: m =1, B m = 2) stick together and mass becomes A+B = M = 3
Suppose that $v_a = 6m/s$ and $v_b = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6$$v_b = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6, v_{cm} = p/M = 2$
After collision velocity would be anyway lower as KE should be distributed among more mass, but some KE is lost in the crash. How much?
Momentum is conserved: $ p_{ab} = 6$ , from this datum you can calculate its velocity: $$v_{ab} = v_{cm}= \frac{6}{3} = 2$$ and $E_k = 0.5 * 2^2 *3 = 6 \rightarrow E_a = 2 + E_b = 4$.
Some energy has been transferred to B, but two thirds of the energy have been lost.
Velocity of center of mass is the same, although KE has changed.
Please note that momentum is conserved because we are assuming that on the surface of contact there is no friction.
...how can there be a change in kinetic energy of the system if there is no change in momentum?
A change of KE without a change of momentum is not only possible but very frequent, because as you noted p = mv momentum varies linearly and KE quadratically. You can get the same product by a wide range of factors: 6 = 61, = 32, = 23, = 16, = 0.5*12, etc., different factors give same momentum
All these factors give same values for m*v, but as the figure for v must be squared, you get all different values between momentum and energy, therefore the same factors give momentum = 6, but KE =3, =6, =9, =18, =72, etc, same momentum corresponds to many different values of KE
I hope this clarified all your doubts