Endomorphism rings of isogeneous elliptic curves Let $E$ and $E'$ be isogenous elliptic curves and $K=\text{end}(E) \otimes \mathbb(Q) $
Is it true that $\text{end}(E') $ is a subring of $K$? 
The only thing I thought is that the isogeny between $E$ and $E'$ and its dual give a map between the endomorphism rings, which as far as i know needs not to be surjective or injective.   
 A: Try this:
Let $\varphi:E\to E'$ and $\psi:E'\to E$ be the pair of isogenies, $\deg\varphi=\deg\psi=n$ and $\psi\circ\varphi=[n]_E$, while $\varphi\circ\psi=[n]_{E'}$. I’m going to map $\,f\in\text{End}(E)$ to an element $\,\tilde f\in\text{End}(E')\otimes\Bbb Q$, and we’ll see that the map is injective.
Simply set $\,f'=\frac1n\varphi\circ f\circ\psi$. You see immediately that it’s a ring homomorphism, and certainly injective, since the composition of (nonzero) isogenies is still nonzero.
I think this gets you to come down where you want to be.
A: I think it is not true at least over $\mathbb{C}$
Let us consider the elliptic curve $E$ whose lattice is given by $\langle 1,\frac{i}{2} \rangle$ which has not the CM
Now let us consider the isogeny given by the quotient for the translation of the point $\frac{1}{2}$ then you get an elliptic curve $E’$ whose lattice is $\langle\frac{1}{2},\frac{i}{2}\rangle$ which is isomorphic to $\langle 1,i\rangle$. Thus $E’\simeq E_i$ which has the CM given by $i$.
This i think that  it may happen then after an isogeny you get an automorphism group bigger or not depending on the isogeny you are considering.
