Constructing $J_x$ and $J_y$ from $J_z$ $\newcommand{\ket}[1]{\left|#1\right>}$
Normally the matrix representation of $J_x$ and $J_y$ are derived using the ladder operators. But is a better intuition possible by starting with $J_z$ alone ? One just needs to find the eigenvectors of these operators in the basis of eigenvectors of $J_z$, but how do you do that ?
For $j$ = 1, the basis states are $\ket{j=1, m_z = +1}$, $\ket{j=1, m_z = 0}$ and $\ket{j=1, m_z = -1}$
if $\ket{j, m_y}$ are the eigenstates of $J_y$ then the basis just as above would be $\ket{j=1, m_y = +1}$, $\ket{j=1, m_y = 0}$ and $\ket{j=1, m_y = -1}$
To represent say $\ket{1, m_y = +1}$ in $J_z$ basis : $$\ket{1, m_y = +1} = \langle{1, m_z = +1}\ket{1, m_y = +1} \ket{1, m_z = +1} + ...$$ and so on for all the three $J_z$ eigenvectors
But how do you intuitively figure out those inner products. For $j = 1$, the momentum vector $\vec{J}$ in real space looks like this:

Then what is the intuition behind $\langle{1, m_z = +1}\ket{1, m_y = +1}$ ?
 A: $\newcommand{\ket}[1]{\left|#1\right>}$

But is a better intuition possible by starting with $J_z$ alone ?

I doubt that. There would be an infinity of solutions, as any rotation between  and  would  provide an equally acceptable set of such. In any case, I'll just provide the conventional ladder operator stick-in-the mud answers for $j$ = 1.
In the same z-basis you started with,
$$\ket{j=1, m_z = +1} = (1,0,0)^T,\\
\ket{j=1, m_z = 0}= (0,1,0)^T,\\
 \ket{j=1, m_z = -1}=(0,0,1)^T,
$$
you have the obvious
$$
J_z=\operatorname{diag}(1,0,-1), \qquad J_y= \frac{i}{\sqrt 2} \begin{bmatrix}0&-1&0\\ 1&0& -1\\ 0&1&0   \end{bmatrix},
$$
$=(J_+-J_-)/2i$, (which you might also obtain by blood-sweat and tears through a π/2 rotation around x if you had the rotation matrix for that, and converted all in the spherical basis... don't go there.)
It is then evident by inspection of the eigenvectors of $J_y$ that
$$\ket{j=1, m_y = +1} = (1,i\sqrt{2},-1)^T/2,\\
\ket{j=1, m_y = 0}= (1,0,1)^T/\sqrt{2},\\
 \ket{j=1, m_y = -1}=(1,-i\sqrt{2},-1)^T/2. 
$$
Do not neglect the requisite conjugation in the dot products of the spherical basis.
