capacitor ripple current in buck converter
I'm not sure I'm following your line of thought, but it sounds like you are attacking the problem backwards from the usual way.

Normally, you'll first set a limit on ripple current (\$\Delta{}I_L\$), and choose an inductor to meet that requirement.

Then you will choose your output capacitor to achieve the ripple voltage that you want.

For idealized components, you will have

$$\Delta{}V_{out} = \Delta{}I_L \sqrt{R_{ESR}^2 + \left(\frac{1}{8 f_{sw} C_{out}}\right)^2}$$

So the lower the ESR and the higher the capacitance, the lower the ripple voltage will be. 

This ripple voltage is ripple due the inherent transients in each cycle of the switching circuit, when operating with a constant load current.

Ripple or ring due to load transients is more due to the voltage control loop than due to these considerations. So to minimize that you need to model your control loop and ensure you have sufficient phase margin to avoid an unwanted ring response to load transients.


  Is this amount of current (ripple current, Irms) identical with Inductor ripple current of regulator?


Yes. If the load current is constant (or near enough) then the deviations of inductor current from the average value (which is the load current) have nowhere to go but in and out of the output capacitor.