rewinding AC induction motor to work with lower supply voltage
To change voltages on a motor but keep the same performance, the number of turns and the wire cross-sectional area have to change proportionally to the ratio of the 2 voltages.  If you want to increase voltage, the number of turns increases and the cross-sectional area of the wire decreases.   If you want to decrease voltage, the number of turns decreases and cross-sectional area of the wire increases.  

One hint to help with this is that in order to double the cross-sectional area of standard magnet wire, you need to decrease the wire a gauge by 3.  So, for example, to double the cross-sectional area of 25 AWG wire, you need 22 AWG wire.  

Also note that doubling the cross-sectional area of the wire decreases the resistance by half, so if you reduce the number of turns by half and double the cross-sectional area, your total resistance will be 1/4th the original resistance.  

For your problem, if you are trying to convert a 230 V motor to a 12 V motor you would need to divide the number of turns by \$230/12 = 19.2\$ and multiply the cross-sectional area of each turn by 19.2.  If your motor has 95 turns and 25 AWG wire, then your 12 V winding would have 5 turns and 12 AWG or 13 AWG wire.  

If there is a capacitor for this motor, it will change inversely proportional to the voltage ratio squared.  So in your case, the capacitor will have to be \$19.2^2 = 368\$ times bigger.