There's an easy way to do that without having to dig into Maths.
What's a geosynchronous orbit ? It's a near-circular orbit, with an equatorial inclination near 0°, and with T = sidereal day ( 86164 seconds).
Geos means "Earth", in ancient Greek.
So, a "marsynchronous" orbit would use the same principles. For that, we need to know the duration of a sideral day on Mars. Thanks Wikipedia :
"The average length of a Martian [ame="http://en.wikipedia.org/wiki/Sidereal_day"]sidereal day[/ame] is 24h 37m 22.663s"
23 seconds will be ok, we don't need that much precision !
24*3600+37*60+23 = 88643 seconds
Now search for a circular (Ecc ~= 0), equatorial (Inc(ref EQU) ~= 0) orbit around Mars with T = 88643 sec.
That should work, unless I missed something obvious, which is possible :lol:
Edit : obviously, I didn't read carefully enough the topic, which was about Sun-synchronous orbit :beathead:
Edit2 : replace "marsynchronous" by "areosynchronous". It makes sense : Ares was the Greek god of War (Mars was the Roman one).