Found a solution requiring
First, label the coins A,B,C,D,E,F,G,H,I,J. Then for the first 4 weighings, weigh:
A/B, C/D, E/F, and G/H (slash notation: "A/B" means "weigh A vs B"). After this, you will have 0, 1, 2, or 3 weighings where the scales differed. Handle each case.
Zero imbalances: This can only happen if one of the weighings had a heavy on each side, and one of the unweighed coins is light. These enumerate (labeling the coins heavy-heavy-light) as: ABI, ABJ, CDI, CDJ, EFI, EFJ, GHI, GHJ. Weigh AI/CJ. If this balances, it's either ABI or CDJ. Weigh A/C to determine which. If AI > CJ, then the options are ABJ, EFJ, or GHJ. Weigh E/G to determine which. Likewise, if AI < CJ, the options are CDI, EFI, or GHI. Again, weigh E/G to determine which.
For (Update: fixed mistake, thanks to comments):
One imbalance: relabel the coins so A>B. The options now are AIJ, AJI, AIB, AJB, CDB, EFB, GHB, IJB. Weigh CI/EJ. If CI=EJ, then the options are GHB, IJB. Weigh G/I to determine which. If CI>EJ, the options are AIB, CDB, or AIJ. Weigh AB/CE to determine which (AIB will balance, the others will lean toward the heavy one). Likewise, if CI < EJ, the options are AJB, EFB, or AJI. Again, weigh AB/CE to determine which.
Two imbalances: relabel the coins so A>B, C>D. The options are now ACB, ACD, ACI, ACJ, AID, AJD, CIB, CJB. Weigh I/J and then AB/CD to determine which of the 8 options is the solution.
Three imbalances: relabel the coins so A>B, C>D, E>F. The only options are ACF, AED, or CEB. Weigh A/C to determine which.
In any case,
at most 6 weighings are required.
And as has been pointed out, there are 360 initial configurations, and $360 > 243 = 3^5$, so at least 6 weighings are required.