Rotating Back Home

This is not a complete solution, as it does not show that $6$ rotations is the minimum, but it does show that $6$ is possible:

We can show that $6$ rotations is possible without having to find a specific angle by seeing that there exists an intersection point between arcs along the surface of the sphere that are the locus of different combinations of $3$ rotations. For example, in the picture from @Michael Mendrin 's solution, the third transformed point would be the intersection of the two arcs that are locus of points for rotations $(y^+, x^+, y^+)$ and $(z^-, x^-, z^-)$ (where $x^+$ is a rotation around the $x$ -axis one way, and $x^-$ is a rotation around the $x$ -axis the other way, and so on) that start at $(1, 0, 0)$ for any angle $\theta < 90°$ . Reversing $(z^-, x^-, z^-)$ gives $(z^+, x^+, z^+)$ , and (making sure that it follows the given requirements) this gives the combined solution of $(y^+, x^+, y^+, z^+, x^+, z^+)$ , which has $\boxed{6}$ rotations.

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This is not a minimum solution, but an interesting asymmetric solution involving the golden ratio:

One possible way to get the point back home with $10$ rotations is $(y^+, x^+, y^+, z^+, x^-, y^-, z^-, y^-, x^-, z^+)$ (where $x^+$ is a rotation around the $x$ -axis one way, and $x^-$ is a rotation around the $x$ -axis the other way, and so on), at an angle of $\theta = \sin^{-1} \sqrt{\phi - 1} = \cos^{-1} (\phi - 1) \approx 51.827°$ , where $\phi$ is the golden ratio equal to $\frac{1 + \sqrt{5}}{2}$ .

Here is a picture courtesy of @Michael Mendrin :

I don't really have a proof for this, other than I noticed that this angle would be a likely candidate as $4$ rotations brings the point to a different axis (for example, $(y^+, x^+, y^+, x^+)$ to $(0, 1, 0)$ ), and in my attempt to use a series of these $4$ rotations in a combination to jump from one axis to another to get back to $(1, 0, 0)$ , I found (with the help of a computer program) that $10$ rotations was the minimum number needed with the given angle $\theta = \cos^{-1} (\phi - 1)$ . Perhaps someone can explain how this works out?

Here's one solution with six rotations:

The exact angle here is $\theta = ArcTan\left( \dfrac { 1 }{ 9 } \left( -2+\sqrt [ 3 ]{ 19-3\sqrt { 33 } } +\sqrt [ 3 ]{ 19+3\sqrt { 33 } } \right) \left( 1+\sqrt [ 3 ]{ 19-3\sqrt { 33 } } +\sqrt [ 3 ]{ 19+3\sqrt { 33 } } \right) \right) =0.99597...$
or approximately $57.0649$ degrees.

The sequence of rotations is $(y^+, x^+, y^+, z^+, x^+, z^+)$

The following are the rotation matrices for rotation about axes $x, y, z$ respectively:

$\left( \begin{matrix} 1 & 0 & 0 \\ 0 & Cos(\theta ) & Sin(\theta ) \\ 0 & -Sin(\theta ) & Cos(\theta ) \end{matrix} \right)$
$\left( \begin{matrix} Cos(\theta ) & 0 & Sin(\theta ) \\ 0 & 1 & 0 \\ -Sin(\theta ) & 0 & Cos(\theta ) \end{matrix} \right)$
$\left( \begin{matrix} Cos(\theta ) & Sin(\theta ) & 0 \\ -Sin(\theta ) & Cos(\theta ) & 0 \\ 0 & 0 & 1 \end{matrix} \right)$

David Vreken has an extremely interesting solution that isn't symmetric, involving a very special angle. I wonder if there are any shorter paths that doesn't involve symmetry.

I found a flaw in my reasoning... will correct it shortly... sorry