If $\phi$ is an endomorphism of an elliptic curve, and $\phi = \hat{\phi}$ then $\phi = [m]$? I heard a reference to this fact, but I cannot find a reference. (I can find the converse in Silverman, namely that $\hat{[m]} = [m]$.)
Notation: $[m]$ is multiplication by $m$ in the group law, and $\hat{\phi}$ is the dual endomorphism of $\phi$.
Question: If $\phi$ is an endomorphism of an elliptic curve, and $\phi = \hat{\phi}$ then $\phi = [m]$?

I think that if $E$ has CM this is false: take an endomorphism $\phi$ such that $\phi^2=[d]$, where $d$ is square-free. Then $\phi\circ\widehat{\phi}=[\deg\phi]=[d]$. Thus $\phi^2=\phi\circ\widehat{\phi}$, which implies $\phi=\widehat{\phi}$, but $\phi$ cannot be multiplication by an integer since it has non-square degree.
