Is a convervative and spherical symmetric vector field a radial vector field? In electrostatic, is well know that the electric field generated by point charged particle $q$ is given by:
$$\vec E(\vec r)=\kappa\frac{q}{||\vec r||^3}\vec r$$
This vector field is clearly conservative. This fact follows from its spherical symmetry (that is, $\vec E(\vec r)=\vec E(A\vec r)$ for any orthogonal matrix $A$) and its radial character ($\vec E(\vec r)=f(||\vec r||)\vec r$, for some function $f$).
My question is: If $\vec E$ is a conservative vector field with spherical symmetry, then $\vec E$ must be a radial vector field
$$\vec E(\vec r)=f(||\vec r||)\vec r \quad ?$$
 A: Note: I'm assuming we're working in $\mathbb R^3,$ since the electrostatic field is a three-dimensional field.
The following answer is written under that assumption, and does not necessarily work in spaces of other dimensions. In particular, if we were working in two dimensions then "spherically symmetric" is just "circularly symmetric" and the arguments below do not hold.

Never mind the conservative part. How can a vector field be spherically symmetric?
"Spherically symmetric" means that I can rotate the vector field around its center in any way I like and I will still have exactly the same vector field, exactly the same as if it had not rotated at all.
Now let's look at the field vector at some point at displacement $\vec r$ from the center of the spherical vector field,
and consider rotations of the spherical vector field around the axis through the origin and the chosen point, that is, rotations parallel to $\vec r$.
The field vector at that point must be unchanged by any such rotation.
What kind of vector can satisfy that property?
If the field vector is not parallel to $\vec r$ then a $180$-degree rotation (for example) will cause the field vector to point in a different direction.
The only way to keep the same direction is if the field vector is
parallel to $\vec r,$ that is, the field vector is a radial vector.
So there you have it. A spherically symmetric vector field is a radial vector field.

More formally, working from the definition that requires
$A\vec E(\vec r) = \vec E(A\vec r)$ for every orthogonal matrix $A$:
For a particular arbitrary displacement vector $\vec r,$
let $A$ be the matrix that performs a $180$-degree rotation about the axis parallel to $\vec r.$
Then $A\vec r = \vec r,$ hence $\vec E(A\vec r) = \vec E(\vec r).$
Spherical symmetry then requires that $A\vec E(\vec r) = \vec E(\vec r)$
for this particular choice of $A$ and $\vec r.$
Let $\hat r = \vec r/\lVert\vec r\rVert,$ that is, $\hat r$ is a unit vector parallel to $\vec r.$ Let $(\hat r, \hat u, \hat v)$ be an orthonormal basis.
Then $\vec E(\vec r) = x_1 \hat r + x_2\hat u + x_3\hat v$ for some real
$x_1,x_2,x_3.$ Since $A$ is a $180$-degree rotation about $\vec r$
(that is, about $\hat r,$)
$$ A\vec E(\vec r) = x_1 \hat r - x_2\hat u - x_3\hat v. $$
We conclude that
$$x_1 \hat r - x_2\hat u - x_3\hat v = x_1 \hat r + x_2\hat u + x_3\hat v,$$
which implies $-x_2 = x_2$ and $-x_3 = x_3,$
which is possible only if $x_2 = x_3 = 0.$
Therefore $\vec E(\vec r) = x_1\hat r,$ that is,
$\vec E(\vec r)$ is a radial vector.
Since the original choice of $\vec r$ was arbitrary,
the vector field $\vec E$ is a radial vector field.

Next we can compare the field vectors at two different points that are each at the same radial distance from the center of the vector field.
We rotate the vector field so that the first point is moved to the place where the second point was. This must not change the field vector. That tells us the field vectors had to have equal magnitude.
Since this is true for any two points at the same radial distance, all points at a given radial distance $r$ from the center of the vector field have field vectors of the same magnitude.
The radial distance $r$ determines the magnitude uniquely.
So now we know that a spherically symmetric vector field is a radial vector field in which the magnitude of the field vector at any point is a function of the radial distance $r$ from the center to that point.
From this it can be shown that the field is conservative.
That is, the phrase "conservative and spherically symmetric vector field" is redundant. If the vector field is spherically symmetric, we already know it is conservative without having to be told.
