So according to what I've read, in a common emitter BJT amplifier, an unbypassed emitter resistor acts as a current-series feedback path. The gain without feedback is: $$A=\frac{I_o}{V_i}=\frac{-h_{fe}}{h_{ie}+R_E}$$ and the feedback factor is: $$\beta=-R_E$$ The output impedance should then be: $$Z_{of}=Z_o(1+A\beta)=Z_o(1+\frac{R_Eh_{fe}}{h_{ie}+R_E})$$ This is a different result compared to finding the Thevinin equivalent resistance looking back into the output which just gives the output impedance as \$R_C\$ whether or not an emitter resistor is bypassed or not.
Any insight on this seeming inconsistency would be helpful. Thanks.