Remove R5 and you will have what you describe. The configuration of Q5 is called common collector or emitter follower. Essentially, the voltage at the emitter is the voltage at the base minus 0.6V, but the emitter current can be much more than the base current, because the gain of the transistor will draw more current from the collector. Thus, it's a current amplifier.
Remember, the base-emitter junction is a diode. So, the emitter will be about 0.6V below the base if you forward bias it. With R5 removed, you can pull the emitter up to \$V_{cc} - 0.6V\$. With R5 present, you won't get it as high, since some voltage will be dropped when current flows in R5.
Since there are things that will limit the current in the emitter leg of Q5, you don't need R5 to limit the base current, which isn't true of Q2 or Q4, which have their emitters shorted to ground, or Q1, with its emitter shorted to \$V_{cc}\$.
See Why would one drive LEDs with a common emitter?
There isn't much difference in performance. In circuit 1, the anode of D1 will be at \$V_{cc} - 0.2V\$, whereas in circuit 2, it will be at \$V_{cc} - 0.6V\$, so the LED current is a bit higher in circuit 1, assuming R1 and R4 are the same value.
Circuit 2 has the advantage that the base current goes towards powering the LED, but since the base current is small, this isn't a big effect.
The last subtle difference is that in circuit 1, Q1 enters saturation, which will charge the base-emitter capacitance. When you then turn it off, this capacitance has to discharge before Q1 really goes off, adding a bit of delay from when your MCU output goes low to when the diode gets switched off by Q1. Q5 never enters saturation, because the emitter voltage is brought up to just the point where the transistor enters saturation, but not more. So, no turn-off delay. The delay is very short, and probably not significant until you are switching at least 50kHz.
Either the MOSFET has an internal gate-source short or it is wired up wrong (or, see Joe Hass's answer in case you're actually applying +5 directly to the base).
If you want to be able to apply +5 without the 1K resistor you show, you can use an N-channel MOSFET rather than a BJT.
Best Answer
You're very close. instead of having the NPN in its present location, it need to be between the motor and GND. So disconnect the NPN's emitter from the motor, its collector from the 9 V supply, and connect: Emitter -> GND Collector -> Motor Base -> 1k & 5 V as you have it.
The motor will go between the 9 V supply and the NPN's collector.
Whether or not 1k will work depends on the motor current. 5 V and 1k will give about 4.3 mA of base current -- probably about 100 mA of collector current. This sounds low for even a small motor.
simulate this circuit – Schematic created using CircuitLab
Also, it is possibly you have (partly) damaged your NPN -- the way you have the circuit now, the base-emitter junction is reverse biased. These usually break down around 6 V, and when they do, the beta (gain) of the transistor gets degraded.