To answer this question you need to consider two things:
- will the transistor have the same biasing ?
- will the small-signal transfer change ?
Let's make things easy on ourselves and assume the beta of the transistor is very high so we can ignore the base current. Then the base voltage would be the same in both circuits. The rest of the circuit (V2,Q1, R2) is identical so YES, the transistor will have the same biasing.
This means that the small signal parameters of the transistors will also be the same !
But will the overall small signal equivalent change ? We only need to consider how the signal comes from V1 into the base of Q1 compared to V3 into the base of Q2. Now I do see a difference ! Do you see it ?
R1 // R3 is indeed 333.3 ohms (= R4). BUT while the signal from V1 is attenuated by R1 and R3, the signal from V3 is NOT !
So the circuit with Q2 will have a higher voltage gain !
Even though from a DC point of view, the transistor cannot tell the difference if it was in either circuit.
This is assuming the input signal has the same amplitude on V1 and V3. But V3 is 10 V DC instead of 15 V DC. If you would also lower the signal amplitude of V3 with the same ratio then the signal at the output would be the same !
Then the circuits would indeed be equivalent !!!
Do you also see that it is not so much the non-linear component that makes the difference ? I mean, the transistor is biased in the same way so it behaves identical in both circuits. But the transfer to the Thevenin equivalent did change the way the signal is attenuated through the circuit.
I do not think you can state that in general it is safe to use the Thevenin equivalent. You have to look at it case by case. And use common sense !