How to drive power MOSFET by logic level signal without inverting phase?
If you want the load to be off when the micro is not powered, the polarity from the driving signal to the output plays no role. When the micro is turned off, its outputs are high impedance, i.e. they act like tiny capacitors with one terminal to ground. Any tiny, little noise can charge these caps, so the voltage on the pins can pretty much assume any value.

You can solve this problem in many ways. Since you use the level shifter I will assume the arduino pins are not 12V tolerant.

Here's your schematic:







simulate this circuit &ndash; Schematic created using CircuitLab

That's actually pretty near to what you have. I am not really sure of why you had all these resistors spread around, anyway let's see what happens in three cases.

vin = 5V
Q3 is on, Q2 is off, mos gate is tied high via R4, load is on

vin = 0V
Q3 is off, Q2 is on, mos gate is tied low, load is off

vin = hiz
this means vin is disconnected. R6 pulls the first bjt base low, turning it off, Q2 is on, load is off.

Please note that this circuit does not use the 5V rail, so if the 5V rail is removed the load is switched off.

If you can be sure that the 5V will always be present when the 12V is present, then you can probably save one transistor.