The states $|+\rangle, |-\rangle$ can be mapped to $|0\rangle, |1\rangle$ by a simple rotation.
But if I now have other states ($|\psi_0\rangle, |\psi_1\rangle$) which are not orthogonal, does a unitary transformation of these states to two other states ($|\phi_0\rangle, |\phi_1\rangle$) exist?
And how can I find this unitary transformation?
There is a unitary that maps $\{|\psi_1\rangle,|\psi_2\rangle\}$ to $\{|\phi_1\rangle,|\phi_2\rangle\}$ if and only if $$ \langle\psi_1|\psi_2\rangle=\langle\phi_1|\phi_2\rangle. $$
Thy "only if" direction is straightforward. Let's assume $U|\psi_i\rangle=|\phi_i\rangle$. Then $$ \langle\phi_1|\phi_2\rangle=\langle\psi_1|U^\dagger U|\psi_2\rangle=\langle\psi_1|\psi_2\rangle. $$
The "if" direction requires a little more care. Let $|\psi^\perp\rangle$ be the component of $|\psi_2\rangle$ that is orthogonal to $|\psi_1\rangle$. Similarly for $|\phi^\perp\rangle$. So, we claim that if we can find a unitary $U$ such that $$ U|\psi_1\rangle=|\phi_1\rangle,\qquad U|\psi^\perp\rangle=|\phi^\perp\rangle, $$ then we are done because $$ U|\psi_2\rangle=U(\cos\theta|\psi_1\rangle+\sin\theta|\psi^\perp\rangle)=(\cos\theta|\phi_1\rangle+\sin\theta|\phi^\perp\rangle)=|\psi_2\rangle $$ where $\cos\theta=|\langle\psi_1|\psi_2\rangle|=|\langle\phi_1|\phi_2\rangle|$.
Now let $\{|\gamma_i\rangle\}$ be any orthonormal basis with $|\gamma_1\rangle=|\psi_1\rangle$ and $|\gamma_2\rangle=|\psi^\perp\rangle$ and let $\{|\lambda_i\rangle\}$ be an orthonormal basis with $|\lambda_1\rangle=|\psi_1\rangle$ and $|\lambda_2\rangle=|\psi^\perp\rangle$. We can define $$ U=\sum_i|\lambda_i\rangle\langle \gamma_i|. $$ This certainly provides the required transformation.
A small rewarding of the argument in this other answer:
Given $\{|\psi_1\rangle,|\psi_2\rangle\}$ and $\{|\phi_1\rangle,|\phi_2\rangle\}$ such that $\langle\psi_1|\psi_2\rangle=\langle\phi_1|\phi_2\rangle$, we can always write $$|\psi_2\rangle = \cos\theta |\psi_1\rangle + \sin\theta|\psi_1^\perp\rangle,\\ |\phi_2\rangle = \cos\theta |\phi_1\rangle + \sin\theta|\phi_1^\perp\rangle,$$ for some angle $\theta$ given by $\langle\psi_1|\psi_2\rangle=\cos\theta$, and for some states $|\psi_1^\perp\rangle,|\phi_1^\perp\rangle$. These are defined as $$|\psi_1^\perp\rangle\equiv \frac{|\psi_2\rangle-\cos\theta|\psi_1\rangle}{\sin\theta}, \qquad |\phi_1^\perp\rangle\equiv \frac{|\phi_2\rangle-\cos\theta|\phi_1\rangle}{\sin\theta},$$ and are such that $\langle\psi_1^\perp|\psi_1\rangle=\langle\phi_1^\perp|\phi_1\rangle=0$.
Given any pair of states $|\psi\rangle,|\phi\rangle$, you can always find unitary operators $U$ such that $U|\psi\rangle=|\phi\rangle$. The choice of $U$ is not unique. Let us then assume that $U|\psi_1\rangle=|\phi_1\rangle$. Our goal will be to show that we can find $U$ such that we also have $U|\psi_2\rangle=|\phi_2\rangle$. But given our observation in the first point, this is equivalent to finding $U$ such that $U|\psi_1^\perp\rangle=U|\phi_1^\perp\rangle$.
We thus reduced the problem to finding a unitary $U$ which maps $|\psi_1\rangle\mapsto |\phi_1\rangle$ and $|\psi_1^\perp\rangle\mapsto |\phi_1^\perp\rangle$. But finding a unitary sending orthogonal states into orthogonal states is trivial to do: for example, we could just define $U$ as $$U = |\phi_1\rangle\!\langle\psi_1| + |\phi_1^\perp\rangle\!\langle\psi_1^\perp|,$$ and this can be readily seen to be unitary and implement the correct mapping.
From your question it appears you know the two sets of states ahead of time. If so and the $|\psi_0\rangle$ and $|\psi_1\rangle$ are not orthogonal, then a unitary transformation that transforms |$\psi_0\rangle$ to $|\phi_0\rangle$ is not guaranteed to transform |$\psi_1\rangle$ to $|\phi_1\rangle$. There could be one, in special cases, but not always.
However, if both sets are orthogonal, then I expect to have a unitary transformation that could do that every time.
Edit: If they are not orthogonal, but the inner products of both the sets are the same i.e. if $\langle \psi_0 | \psi_1 \rangle$ = $\langle \phi_0 | \phi_1 \rangle$ and if $U$ exists where $\phi_0 = U |\psi_0 \rangle$, then it follows $\phi_1 = U |\psi_1 \rangle$.
This is because $\langle \phi_0 | \phi_1 \rangle$ = $\langle \psi_0 | U^{\dagger} U |\psi_1 \rangle$ , and since $U^{\dagger} U = I$ .
And $U = \sum_{j} |\phi_j \rangle \langle \psi_j |$
Ref: Qiskit Universality
