I have asked this question before and, while the accepted answer solved my problem back then, I am still interested in finding the rotation axis and angle. Let me rephrase the problem here:
I would like to find the rotation axis and angle needed to align two 3D vetor bases, $(\vec a, \vec b, \vec c)$ towards $(\vec d, \vec e, \vec f)$ (meaning that, after the rotation is applied, $(\vec a, \vec b, \vec c)$ equals $(\vec d, \vec e, \vec f)$). These bases are orthonormal (i.e. normalized vectors, forming 90-degree angles between any of them). The bases also have the same handedness (i.e. no mirroring is necessary to align one towards the other).
Now, due to these bases having the same handedness, there must exist a single rotation that aligns one towards the other (right?).
I would like to know the solution out of curiosity. There's no concern for stability/precision. I've tried solving the original question's accepted answer symbolically, then extracting the axis and angle from the matrix -- did not go so well. Maybe finding the eigenvector would help?
The rotation matrix, $R$, can be found relatively easy, such that,
$$ R\,\vec{a} = \vec{d}; \quad R\,\vec{b} = \vec{e}; \quad R\,\vec{c} = \vec{f}. $$
The rotation matrix/second-order-tensor can be constructed by adding three dyads,
$$ R = \vec{d}\vec{a} + \vec{e}\vec{b} + \vec{f}\vec{c}. $$
When using this notation the multiplication with a vector should be written as a dot product. From this it can also be proven that this expression for $R$ is correct. For example with mapping $\vec{a}$ to $\vec{d}$, the first dyad will return $\vec{d}\vec{a}\cdot\vec{a}$ and since $\vec{a}$ is normalized that is equal to $\vec{d}$. The remaining dot products will yield zero, because the other vectors on the right side of each dyad is orthogonal to $\vec{a}$.
Each dyad can also be written as a 3$\times$3 matrix using the vector direct product. The total 3$\times$3 rotation matrix can then be constructed by adding the three 3$\times$3 matrix from the vector direct products.
The axis of rotation is equal to the eigenvector of the eigenvalue equal to one. For finding this eigenvector you can take a look at the answer to this question. In order to find the axis of rotation you have to calculate the following matrix,
$$ R-R^T = \begin{bmatrix} 0 & \alpha & \beta \\ -\alpha & 0 & \gamma \\ -\beta & -\gamma & 0 \end{bmatrix}, $$
such that the normalized axis of rotation, $\vec{r}$, is equal to,
$$ \vec{r} = \frac{ \begin{bmatrix} -\gamma & \beta & -\alpha \end{bmatrix}^T }{\sqrt{\alpha^2 + \beta^2 + \gamma^2}}. $$
The angle of rotation, $\theta$, can be found according to,
$$ \text{Tr}(R) = 1 + 2\cos\theta, $$
where $\text{Tr}(R)$ is the trace of $R$, the sum of its diagonal elements. Due to the cosine you do have to look at the sign of the angle and direction of $\vec{r}$.