Performing EdDSA/Ed448 employing Montgomery ladder
According to Alternative Elliptic Curve Representations,  each other point $(x1,y1)$ of Edwards448 corresponds to the point $(u,v)$ of Curve448, where:
$u = y1^2/x1^2$
$v = y1*(2-x1^2-y1^2)/x1^3.$
Under this isogenous mapping, the base point $(G1x, G1y)$ of Edwards448  corresponds to the base point $(Gu,Gv)$ of Curve448. The dual isogeny maps both the point at infinity and the point $(0,0)$ of order two of Curve448 to the point $(0,1)$ of Edwards448, while each other point $(u,v)$ of Curve448 corresponds to the point $(x1,y1)$ of Edwards448, where:
$x1 = 4*(u^2-1)*v/((u^2-1)^2+4*v^2)$
$y1 = u*((u^2-1)^2-4*v^2)/(2*(u^2+1)*v^2-u*(u^2-1)^2).$
under this dual isogenous mapping, the base point $(Gu, Gv)$ of Curve448 corresponds to a multiple of the base point $(G1x, G1y)$ of Edwards448, where this multiple is $l=4$ (the degree of the isogeny)