Electrical – Mains supply to replace 6 AA batteries

You should use DC-DC converters or dedicated battery packs for every voltage.

You should not use middle points of battery packs. This lead to different charge levels in cells. As a result you'll get an inverse voltage on cells which are discharged first and this will damage the battery and possibly the equipment.

Schematic below shows this.

Inverse voltage is appearing when one of the sells in a battery is completely discharged thus have no voltage across it. If other cells continue to deliver the power, current will still flow thus 'charging' the dead cell with negative voltage.

Load connected to the dead cell this way get reverse polarity supply voltage, thus, possibly, die.

^{simulate this circuit – Schematic created using CircuitLab}

Most optimal (if you do not want to use other voltage rated devices) I think is to supply the most powerful device (6V headlight) directly and use step-up/step-down converters for all others.

That works very well, but do keep in mind that some flash strobes take more than 2A from the battery pack (e.g. my 550EX peaks at about 2.7A when run directly from a bench power supply).

Another option - which I don't know if the 430EX II supports - is to attach a HV power supply that directly charges the flash capacitor, e.g. the Canon CP-E4. This reduces cycle time and is more efficient for the batteries (i.e. you need to take less batteries with you to get the same amount of flashes).