Sure.

  Myf = f(Myf)
  yyf = f(yyf)

Suppose yxf = f(xxf) for all x. Now y is simply

  (lambda x f . f(xxf))

Note that this doesn't get you the applicative-order version we want, but it does satisfy the definition of Y

  My = (lambda x . x x) (lambda x f . f (x x f))
  = (lambda x f . f (x x f)) (lambda x f . f (x x f))
  = (lambda f . f ((lambda a g . g (a a g)) (lambda a g . g (a a g)) f))

etc.