Confused by transistor behaviour
What you are seeing makes sense.  Within nothing connected to the collector, the transistor acts like a diode from B to E.  Current flows out the digital ouput, thru the 1 k&Omega; resistor, thru the transistor from B to E, thru the 470 &Omega; resistor, thru the LED, and back to the digital board.  The LED is dimly lit because it effectively has 1.47 k&Omega; and the B-E voltage drop in series with it and a 5V supply.  That probably results in around 1.5 mA LED current, which is visible but dim.

When 9V is applied to the collector, then the transistor can amplify.  The collector current will be the base current times its gain.  Let's say the LED drops 2 V and the B-E junction 700 mV.  That leaves 2.3 V accross the 470 &Omega; (use designators already!) resistor for about 5 mA LED current.  If the transistor has a gain of 50, then 1 part of this comes from the base and 50 parts of this come from the collector.  The base current is then only about 96 &micro;A, which causes only 96 mV drop accross the base resistor.

There is no need for the base resistor in your setup with the LED and its resistor in the emitter leg.  Just get rid of it and tie the digital output directly to the base.