Powering inverter gate needed for enabling a DC-DC converter
By the way, this is the buck converter you listed, its enable pin supports voltages up the \$V_{IN}\$ value.

I don't know if this is the 'wisest' way, as you call it, but you could do something like this:







simulate this circuit &ndash; Schematic created using CircuitLab

I am assuming the enabling signal would be high when not asserting and low when asserting. If it is open drain, you could pull up the MOSFET's gate anyway—that is why I added the optional pullup. 

So in summary, when the enable signal coming from the main board is low (0V), the MOSFET is OFF and the EN pin on the buck converter side will be pulled high (converter is ON). When the enable signal coming from the main board is high (12V) or open drain, the MOSFET's gate will be high and the MOSFET is ON—this will force a low voltage at the EN pin on the buck converter side, thus disables it (converter is OFF).

You could do a similar thing for the other buck converter. Hope this helps.

Reference: MUN12AD03-SH