Diode-connected transistor, small-signal, Norton, Thevenin
Thevenin and Norton equivalents typically involve independent voltage and/or current source(s). But your only current source here is \$g_m v_{\pi}\$ and it is dependent on \$v_{\pi} = v_{be}\$ (which in this case equals \$v_{ce}\$ because of the diode connection).

To find the equivalent resistance apply a test voltage \$v_x = v_{ce}\$ across C and E, and find the current \$i_x\$ through it. The current is

$$i_x = \frac{v_x}{r_{\pi}} + g_m v_{x}$$

where the first term comes from the current through \$r_{\pi}\$ and the second from the current through the dependent source \$g_m v_{\pi}\$. Also note \$v_{\pi} = v_x\$ (again, the diode connection). Now just solve for \$v_{x}/i_{x}\$.