Let $V$ be a $1$-dimensional $\mathbb{C}$-vector space and $\Lambda \subset V$ be an elliptic curve (=lattice). Let $C : V \rightarrow V$ be the multiplication by $i$. Consider the two following properties.
(a) $\Lambda$ has complex multiplication. That is $\{ A \in \operatorname{End}(\Lambda)|AC=CA\}$ is bigger than $\mathbb{Z}$.
(b) The endomorphism $\operatorname{End}_{\mathbb{R}}(V) \rightarrow \operatorname{End}_{\mathbb{R}}(V), A \mapsto CAC^{-1}$ comes from an endomorpshim $\operatorname{End}_{\mathbb{Q}}(\Lambda \otimes \mathbb{Q}) \rightarrow \operatorname{End}_{\mathbb{Q}}(\Lambda \otimes \mathbb{Q})$.
Does (a) implies (b) ?
EDIT: I will ask my question differently. Suppose that $\Lambda$ has complex multiplication. Denote by $K \subset \operatorname{End}(\Lambda) \otimes \mathbb{Q}$ its endomorphism algebra (which is a quadratic imaginary field). Then we can define the following sub-algebraic group of $GL_W$ where $W= \Lambda \otimes \mathbb{Q}$ :
$$ MT(A) := (K \otimes_\mathbb{Q} A)^\times \subset GL_W(A),$$ for all $\mathbb{Q}$-algebra $A$. On the otherhand, define $G_{\mathbb{R}}$ the sub-algebraic group of $(GL_W)_\mathbb{R}$, where $$G(A) := \{ x \in GL_W(A) : Cx=xC \}.$$
Is $MT$ equal to $G$ over $\mathbb{R}$ ?