Here $j$ stands for the j-invariant of elliptic curve. The case of when the lattice is an integrally closed ring was treated in in "Advanced Topics in the Arithmetic of Elliptic Curves" Chapter II. I'm interested in non-integrally closed case. For example, when $\Lambda=\{\mathbb{Z}+\mathbb{Z}*ni\}$. It hase endomophism ring $\mathbb{Z}[ni]$. So conjugates of $j(ni)$ are (maybe subsets of) j-invariants of elliptic curves with complex multiplication $\mathbb{Z}[ni]$. And seemingly we need to classify ideals of $\mathbb{Z}[ni]$ mod principal ideal. Here is where I'm stuck. Are there reference treating non-integrally closed case? Thanks.
Some update:
A seemingly effective approach is to consider modular equations, which is a polynomial $F$ s.t. $F(j(\tau),j(n\tau))=0$, insert in $j(i)=12^3$ it's possible to obtain a polynomial $f:=F(j(i),-)$, so $j(ni)$ is a root of $f$. Note that $f\in\mathbb{Z}[x]$, has degree $n+1$.
So now we only need to know how $f$ factorizes. And if my calculation is correct, $f$ should has roots $j(ni),j(i/n),j((i+1)/n),\dots,j((i+n-1)/n)$. Note that $j(ni)=j(i/n)$ by modular relations, so $j(ni)$ is a multiple root. And a special case is, if $n$ is prime, the only other multiple roots are $j((i+1)/n)=j((i-1)/n)$, so maybe by some Galois action argument, we have $[\mathbb{Q}(j(ni)):\mathbb{Q}]=2$, is this correct?
If $\mathcal{O}$ is an order of quadratic field $K$ with conductor $f$, then its Picard number $h(\mathcal{O})$ is very simply related to the class number $h$ of $K$. (see here or Primes of form x^2+ny^2 by David Cox)
$$h(\mathcal{O}) = \frac{hf}{[\mathcal{O}_K^\times : \mathcal{O}^\times]} \prod_{p\mid f} \left( 1- \frac{1}{p}\left( \frac{d_K}{p} \right) \right)$$
Standard theory of CM says $[\mathbb{Q}(j(n\sqrt{-1})):\mathbb{Q}]=h(\mathcal{O})$ with $\mathcal{O} = \mathbb{Z}[n\sqrt{-1}]$, so $$[\mathbb{Q}(j(n\sqrt{-1})):\mathbb{Q}] = \frac{n}{a} \prod_{p\mid n} \left( 1- \frac{1}{p}\left( \frac{-4}{p} \right) \right) \qquad a = \begin{cases}1 & n = 1\\ 2 & n\geq 2 \end{cases}$$