AutoHotkey: what script to write to change layout with Fn?

Question

autohotkey

Good afternoon!
Help, please :)
I need to write a code for AutoHotkey so that on the ASUS N46VZ laptop (Windows 10) the Russian-English keyboard layout changes when the Fn button is released.
At the same time, if another button was also pressed while Fn was pressed (ie, Fn was used for a keyboard shortcut, for example, Fn + F6 brightens the screen), then the layout should not change. The behavior should be exactly like the behavior of the Win button - the Start menu only opens when the Win button is released, provided that it was pressed separately from other buttons.

Answer 1

You will have to find the code for the Fn key, but the code that works for me is this:

ih := InputHook("L1")
ih.KeyOpt("{All}", "NV") 
ih.OnKeyDown := Func("OnKey").bind("Down")
ih.OnKeyUp := Func("OnKey").bind("Up")
ih.OnEnd := Func("OnEnd")

~LCtrl::
 ih.Start()
Return
Return

OnKey(direction, ih, vk, sc) {
 if (direction == "Up" and vk == 162) { ;вместо 162 код клавиши Fn
  PostMessage, 0x50,0x2, ,, A
 }
 ih.Stop()
}

OnEnd(ih) {
 ih.Stop()
  
}