In quantum computing, we can always create an arbitrary superposition of states by rotation of $$|0\rangle$$ state for one qubit. This raises a question: for arbitrary superposition of states, is there always a Hermitian operator other than the identity such that this superposition of states is the eigenstate of this Hermitian operator? If we can not find such Hermitian operator, does it mean that such states can not be realized in real life?
Yes. A state $$|\Psi\rangle=|\Phi\rangle$$ is always observable as the non-zero eigenvalue of the Hermitian observable $$\hat O=|\Phi\rangle\langle\Phi|.$$
Furthermore you can explicitly construct the mapping that you are talking about by falling back to the two-level case: define $$|1\rangle = \frac1{1-|\langle 0|\Phi\rangle|^2}\Big( |\Phi\rangle- |0\rangle~\langle 0|\Phi\rangle \Big)$$ Then you have $$|\Phi\rangle$$ living on the $$xz$$-plane of the Bloch sphere and these must be related by some rotation about the $$y$$-axis, $$|\Phi\rangle = \exp\Big[ \theta\big( |0\rangle\langle1|-|1\rangle\langle0| \big) \Big] |0\rangle$$
Given two vectors of the same length, there is always a linear isometry taking one to the other. So let $$w_1$$ be any element of your favorite orthogonal basis $$\{w_i\}$$, find an isometry $$L$$ that takes $$w_1$$ to your favorite state $$s$$, and the image of $$\{w_i\}$$ is the orthogonal basis you''re looking for.
Or: Start with $$s$$, replace your vector space $$V$$ with $$V'=V/$$, find an orthogonal basis for $$V'$$, lift back to $$V$$ and combine with $$s$$ to get the basis you're looking for.