As above, the maximum N-CPC voltage that should be measured at the far end of the circuit, is half the permissible volt drop. 5% x 230V /2 = 5.75V. That is the drop along the length of the neutral conductor (the other 5.75V is in the line). If N & E are at the same potential at the origin, they must be 5.75V apart at the end of the circuit.
But returning to the OP's problem, all the evidence points to the neutral being disconnected at the source end, therefore floating for the whole length of the cable. Because of its length, there is quite a high capacitance between L & N, and between N & E, which can pass enough current to give a steady indication on a voltmeter or MFT. It's an extreme case of 'ghost voltage' if you like.
The big clue is that the voltage L-N and N-CPC add up to 'about 230V' so it is working like a voltage divider, with 147V of the 247V supply across L-N and the remaining 100V across N-CPC. The difference is accounted for by the difference in capacitance. Because the CPC (armour) has a greater surface area facing the floating N conductor than the line, its stray capacitance is higher, its capacitive reactance lower and therefore the neutral will float to a voltage somewhat less than half of the supply, in this case 100V.