True. Let's approach it analytically.
You can calculate the output impedance of that easy enough if you have both divider resistors. Could only see one here so we're sort of pissing in the dark a bit. From that divider and supply you can derive the quiescent current ((Vb-0.6)/150) and therefore small re from Ebers-Moll (26/Ie(mA)). From that you'll find that the emitter can quite happily source a fair bit of current there through small re. But it can't sink it easily as the entire sinking ability is defined by large Re which is 150 ohms here. It's not a push-pull stage. So the amplifier's ability is asymmetrical for low impedances. Really around 400-500 ohms or so is probably the lowest impedance you can chuck on that and get something reasonable looking out.
For me, if it had a common emitter amp on the output with Zin of 1K and out impedance of 1K into a 4:1 matching transformer you'd get a good 50 ohm match cheaply. In true coincidental TV style, here's one I made earlier. Well sort of. A few days back. It's a 7MHz double buffered FET then matching amp. This is a spot on match into 50 ohms confirmed with a return loss bridge (assuming that isn't total poo which it might be

)

Bob Pease style! (this is just a test rig, not the final thing!)
With an inverter you can usually stack them with a suitable series resistance. Say 200 ohms and 4 gates out and that will give you a pretty good 50 ohm match.
Edit: you can sort any matching problem out easily with attenuators though at the cost of amplitude. Stuff a 20dB in line! Problem solved. But that's lazy.
Edit 2: also EMRFD chapter 1. Excellent example of how to build matching networks in here:
http://www.arrl.org/files/file/Product%20Notes/chapter_1.pdf ... see 1.21 (really damn good book!)