How to solve this equation involving an exponential of the angular momentum operator? Let $\vec a$ be a given vector, and  $\vec J=[\hat J_x,\hat J_y,\hat J_z]$ the total angular momentum operator. $\vec \beta$ is an unknown vector, to be solved for, which satisfies
$$
\vec a=\mathrm{Tr}[\vec J\exp(\vec \beta\cdot\vec J)].
$$
How to solve this equation? The above nonlinear equation has three components, and three unkowns, $\vec \beta$.  The existence and uniqueness of the solution are not guaranteed.

I asked several related questions like this, this and this, but I'm still confused  about this problem.
The title and tags of this post are not clear. Thanks for any improvement.
 A: Your stumbling block might be starting from seeking a general solution for something belonging to the universal enveloping algebra, something essentially outside Lie Theory (rotations). You are probably aware of the expansion of your exponential operator in terms of finite polynomials of increasing order with the dimensionality of the representation, so your equation complicates  substantially with increasing this dimensionality.
Instead of rapid-fire XY-problem questions, you might explore your equations for tractable small dimensionalities.
For instance, for spin 1/2, and $\vec \beta\equiv  \beta ~\hat n $, you simply have
$$
e^{\vec \beta\cdot \vec \sigma }= I~\cosh\beta + \hat n \cdot \vec \sigma ~ \sinh \beta \implies \\
2\vec a= \operatorname{Tr} (\vec \sigma ~~ \hat n \cdot \vec \sigma  \sinh \beta)= 2 \hat n  ~ \sinh \beta.
$$
For spin 1, you use the Rodrigues rotation formula,
$$
e^{\vec \beta\cdot \vec \sigma }= I+ \sinh\! \beta~\hat n\cdot J +(\cosh\! \beta -1) ~(\hat n\cdot J),  \implies \\
\vec a = \operatorname{Tr} (\vec J ~~(\sinh\! \beta~\hat n\cdot J +(\cosh\! \beta -1) ~(\hat n\cdot J)^2)),
$$
where you may use the Cayley-Hamilton theorem for 3×3 matrices, etc.
For instance, you may rotate inside the trace $\hat n\cdot\vec J\to J_3$,
$$\vec\beta\cdot \vec a= \beta \sinh\beta \operatorname{Tr} (  J_3^2)= 2 \beta \sinh\beta , $$
since odd powers of $J_3$ are traceless. This component, $\hat n ~~\hat n\cdot \vec a    =\hat n~\vec \beta \cdot \vec a/\beta  $, expands effortlessly to all dimensionalities. Work them out.
But the other two components, $\vec a -\hat n ~ \hat n\cdot \vec a$,  are a messy problem requiring (lacking) knowledge of anticommutators. The problem quickly grows in complication with increasing dimensionality, but there may be an elegant solution shortcut lurking there...
