Convert Ammeter to Voltmeter
You are missing one piece of information - the coil resistance.

Once you have that we can work out the required series resistance, \$ R_M \$.

First calculate the full-scale voltage \$ V_{FS} = {I_{FS}}{R_M} \$ (from Ohm's law).

To convert it to 5 V full scale we need to add a series resistor to drop the excess voltage at 5 mA.

$$ R1 = \frac {5 - V_{FS}}{I_{FS}} = \frac {5 - 0.005R_M}{0.005} \Omega$$

Post a photo of the meter and quote the resistance and report back.



Figure 1. For reference this series from ST have a 196 mV drop at 5 mA. (Photo is a sample from the range.)

For the meter shown in Figure 1 we can calculate the series resistance as

$$ R1 = \frac {5 - V_{FS}}{I_{FS}} = \frac {5 - 0.196}{0.005} = 960 \Omega$$




  This is it opened, the resistance is about 2.9k (if I measured it correctly).


There's something wrong here! To get 5 mA through 2.9 kΩ you need \$ V = IR = 0.005 \times 2900 = 14.5~V \$.

You've either got an incorrect measurement or a voltmeter!