for a current source, the output resistance is the larger the better, and this is still useful to the current mirror. It means that, when the output resistance of your current is large, the voltage at the output node will affect the output current little.
Now, let's back to the second chapter of Razavi's book, and refer to equation (2.38), you will find the output resistance of MOS transistor in the saturation region is r0=1/λID
and than, back to equation (2.26), you will find an assumption above the equation:
ΔL/L=λVds.
In the common process, ΔL is a constant, so when you use large L, than small λ will be got, and causes the large output resistance to improve the stability of the current mirror. So when you use the minimum length of the transistor, you can not copy the current very will, since the node voltage affects the output current hugely.