= Solution
For $a,b>0$, dilation by $t=(b/a)^{1/m}$ maps $X_a$ to $X_b$, because $p(tx)=t^mp(x)=b$. Its inverse is dilation by $t^{-1}$, so \b[all nonempty positive level <manifolds> are diffeomorphic].
A negative level need not be diffeomorphic to a positive one, even when both are nonempty. For $p(x,y,z)=x^2+y^2-z^2$, $X_1$ is connected (parametrize it by $(\sqrt{1+z^2}\cos\theta,\sqrt{1+z^2}\sin\theta,z)$), whereas $X_{-1}$ has two components, according to $z=\pm\sqrt{1+x^2+y^2}$. Thus \b[the answer for opposite signs is no]. If $m$ is odd, the additional symmetry $x\mapsto-x$ does interchange opposite levels up to scaling, but the question allows even degrees.
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