Suppose I have shown for a single vector $\lvert 0 \rangle$ that
$$\langle 0 \rvert U^\dagger \phi U \lvert 0 \rangle=\langle 0 \rvert \phi \lvert 0 \rangle$$
where $\phi$ is a certain operator and $U$ a certain unitary transformation,.
Which properties of $\phi$ would allow me to infer that $U \lvert 0 \rangle=e^{i\alpha}\lvert 0 \rangle$ for some phase $\alpha$? For example, if $\phi$ is the identity then the equation holds for all $\lvert 0 \rangle$ so you cannot deduce anything, but I have the feeling that for certain $\phi$ you can.
If I understand well, you are assuming that the equality holds for all $\phi$.
Then, yes, I think you are right. If you don't mind, I will use slightly different notations: $x = \langle 0 |$. I then take for $\phi$ the orthogonal projector $\pi$ on the line generated by $x$: $\pi(x) = x$, $\pi_{|x^\perp} = 0$.
$$\|x\|^2 = |\langle x, \pi(x)\rangle| = |\langle U(x), \pi(U(x))\rangle| \leq \|U(x)\| \|\pi(U(x)) \| \leq \|x\|^2. $$ So the inequality is an equality. Equality in the Cauchy-Schwarz inequality implies that the vectors are proportional. There is some constant $C$ such that $U(x) = C\pi(U(x))$. Since $\pi(U(x))$ is proportional to $x$, $U(x)$ is proportional to $x$ and we are done.