Let $\mathbf{\sigma}=\sigma_1+\sigma_2+\sigma_3$, where the $\sigma_i$ are the Pauli matrices and define:
$$M:=i\mathbf{\sigma}\cdot \mathbf{v}$$
The claim is that if I change $M$ through a similarity change $M'=SMS^{-1}$, then I can find another vector $\mathbf{v}'$ such that $M'=i\mathbf{\sigma}\cdot \mathbf{v}'$. I checked it by explicitly finding the coefficients $\{f^i_1,f^i_2,f^i_3\}$ for each $\sigma_i$ such that
$$S\sigma_iS^{-1} = \sum_j f^i_j\sigma_j$$
For example, if $S=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, then
$$S\sigma_1S^{-1} = (bd-ac)\sigma_1 - i(bd-ac)\sigma_2+ (ab+bc)\sigma_3.$$
My question is: could I have convinced myself that I can find $\mathbf{v}'$ without actually going through the calculation?