Re-winding an inductor to decrease inductance
Inductance 

Inductance is proportional to turns squared, all things being equal.

L2 / L1 = N2^2 / N1^2 ... 1 so
N2^2 = N1^2 x (L2/l1) ... 2 so
N2 = N1 x sqrt (L2/L1) .... 3  

Apply 3. above

For 15 uH:
L1 = 22 uH
L2 = 15 uH
N1 = 50  so
N2 = 50 x sqrt(15/22) = 50 x 0.826 
=~ 41.3 turns  

So 41 turns in theory and maybe 40 - 42 range.

Similarly for 7 uH N = 50 x sqrt(7/22) =~ 28turns

Current rating:  

Core magnetisation depends on "ampere-turns" so current handling capacity will increase with the inverse of the turns ratios.
I2 = I1 x N1/N2

So 15 uH &amp; 41 turns will handle
9A x 50/41 =~~~= 11A
...
Thicker wire may be required depending on actual currents used, as copper losses increase with I^2.

Treat all such results with care and test in practice.  

Insulation: 

You do not describe the core type or former used.
You say 500V rating - I don't know what voltage you are using in practice but, be sure that insulation is (more than) adequate for the task. As you intend to use existing windings, take care not to nick insulation or have wire rub against core edge. If a bobbin former or insulated core is used this should be easy enough to manage. Odds are a 500-V rated part does not use a bare core and depend on wire insulation alone :-)