For a Y connection, why would a line-neutral current be the same as a line-line current?
In a wye connected 3-phase source, \$I_{Line} =  I_{Phase}\$, because there is only one path.  The current that flows in the line MUST flow in the phase.

There is no such thing as line-to-line current.  



The phase voltages are out of phase by 120\$^{\circ}\$, which means phase currents will be out of phase by 120\$^{\circ}\$ for a balanced load.  

The magnitude will be the same, so \$I_1 = I_2 = I_3\$, but \$ \vec{I_1} \ne \vec{I_2} \ne \vec{I_3}\$, because of the phase angles.

If you do vector addition on these currents, the total is 0, which means there is no neutral current.
$$ \vec{I_N} = \vec{I_1} + \vec{I_2} + \vec{I_3} = 0$$

Because of the 120\$^{\circ}\$ phase shift, ac current will be flowing out and back to the sources at different times.  



When maximum positive current flows out on \$I_2\$ (blue) (as shown in the drawing), \$I_1\$ (red) and \$I_3\$ (green) are 50% negative.  Adding the instantaneous values at any instant in time will give 0.

If the loads are unbalanced, the unbalanced current will flow back to the sources over the neutral wire.

$$ \vec{I_N} = \vec{I_1} + \vec{I_2} + \vec{I_3} \ne 0$$