Current Mirrors Design in IC
Why is the current I4 copied to the transistor M10 (i.e. M10 is a current mirror 1:1)?


Because the \$V_{gs}\$ of M10 is equal to the \$V_{gs}\$ of M9. And if you are operating in the saturation region of the MOSFET, the \$I_d\$ mostly depends on \$V_{gs}\$ and only weakly (due to the channel length modulation effect) on \$V_{ds}\$.


  can you explain me intuitively why the number of transistors is equal to the multiplication factor of the current?


Because the drain-source channels of the N FETs are connected in parallel. And the total current flowing through a parallel combination of parts is equal the sum of the currents through the individual parts.

In comments you added,


  Surely M9 works in saturation because its gate is connected to its drain, but how can we know if also M10 work in saturation? It depends not only on Vgs, but also on Vds, and Vd of M10 depends on the rest of the circuit in A.


Yes, it depends on \$V_{ds}\$. If \$V_{ds}\$ is too low, M10 will enter the triode region and the output current will no longer track the input current. 

And small variations in \$V_{ds}\$ will have small effects on the output current, due to the channel length modulation effect.

This is expected and common to many kinds of current sources. They have a compliance voltage range (a range of output voltages over which they can work properly) and a non-zero output conductance (some variation in output current when output voltage changes).


  Moreover, why don't we connect also M10 as M9 (gate to drain)? 


If we connected M10's drain to its gate, we' also be connecting it to M9's drain  (or rather, M11's drain in this case). We'd be shorting out our control signal to our output current. We'd no longer have isolation between input and output of the controlled current source.