We know that the $j$-invariant of any Supersingular Elliptic Curve (over ${\mathbb{F}_{q = {p^n}}}$) falls on the field ${\mathbb{F}_{{p^2}}}$ . (Silverman-The Arithmetic of Elliptic Curves-Theorem 3.1.a)
My question: Does every vertex of the supersingular isogeny graph $j_0 \in {\mathbb{F}_{{p^2}}}$, have a Montgomery form $${{\rm E}_{\left( A \right)}}:{y^2} = x\left( {{x^2} + Ax + 1} \right)$$ that ${j_{{{\rm E}_{\left( A \right)}}}} = {j_0}$ and $A \in {\mathbb{F}_{{p^2}}}$?
The j-Invariant of ${{\rm E}_{\left( A \right)}}$ is $${j_{{{\rm E}_{\left( A \right)}}}} = \frac{{256{{\left( {{A^2} - 3} \right)}^3}}}{{{A^2} - 4}}$$
I tried this practically with the help of the Magma system, and it happened for many examples; But I do not know the reason for this.