Solution
= Solution
Applying the functional equation at $x^{\alpha^{-1}}$ repeatedly gives
$$
f(x)=f(x^{\alpha^{-1}})=\cdots=f(x^{\alpha^{-n}}).
$$
Since $x^{\alpha^{-n}}\to1$, continuity at $1$ gives $f(x)=f(1)$. Thus \b[$f$ is constant].
Without continuity the claim is false. The function $f(1)=1$ and $f(x)=0$ for $x\ne1$ satisfies $f(x)=f(x^\alpha)$ because $x^\alpha=1$ for positive $x$ exactly when $x=1$, but it is not constant.