Default Action/Effect in an Action Set in PSE5

I have a Photoshop action set that contains a "main" action that performs some functionality and calls a bunch of "sub" actions. The main action is listed first in the set and is automatically picked up as the default action and executes successfully when installed into PSE6, PSE7 & PSE8 and also works as expected in a German version of PSE5 but in an English version of PSE5, somehow the second action (ie a "sub" action) is being executed rather than the first. Is there any way of configuring the default action or forcing PSE5 to recognize the first action as the action to run? If not, does anyone know what algorithm PSE uses to determine the action to run? The action being run isn't first (or last) alphabetically nor positioned first (or last) in the action set. Any thoughts/assistance would be greatly appreciated. Thanks.

Thanks for the response. I think I confused things by bringing up the German version... The action was created in a standard English version of PS CS2 then adjusted and resaved in a standard English version of PS7. It was then packaged and tested in English (and non-English) versions of CS3, CS4, CS5, PSE6, PSE7 & PSE8 without any issues. It was also tested in a German version of PSE5 and worked without any concerns but when tested in 2 different standard English versions of PSE5, it appears that the second action in the action set is the one that is being recognized and run, rather than the first action in the set. The first action is named as "Make Patterned Papers" and the other 20 or so actions are named as "code -- (reference number)" in a lame attempt to slightly obscure the functionality from the PS users. The first "code -- (reference number)" action listed after the "Make Patterned Papers" is the one that is being picked up by PSE5 as the action to run. The actions are not listed in alpha/numeric order and the action that is being picked up by PSE5 is not the first alphabetically either. I tried copying all of the actions to a new action set and using this instead, but had the same result.

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