I am reading Schoof's 1995 paper, Counting points on elliptic curves over finite fields, page 236, Proposition 6.1(i). I am trying to understand page 238 (second paragraph) of the proof: if the conductor is $1$ then $j$ invariant is $0$, or $1728$.
In Silverman's book, The Arithmetic of Elliptic Curves, Section III.10 on page 103, a relation between the Automorphism Group and $j$-invariant is given. Is there any relation between the endomorphism ring of an elliptic curve to the $j$-invariant? When $j=0,1728$, how does the endomorphism ring look?
If the endomorphism ring is equivalent to the maximal order of the field, why is the $j$ invariant equal to $0$ or $1728$?
I know that given $E: y^2=x^3+Ax+B$, if $A=0$ then $j=0$ and if $B=0$ then $j=1728$. The endomorphism ring will be isomorphic to an imaginary quadratic field if $E$ is ordinary and to a quaternion algebra if $E$ is supersingular.
Please help me to understand the relation between the endomorphism ring and $j$ invariant.