First let's see where the equation $\mathbf{E}=-\nabla \phi$ comes from. In a nutshell: you start from the electric field $\mathbf{E}$ which is a modification of space such that charges $q$ experience a force $q\mathbf{E}$. If you move this charge around the field along a path $C$ you change its potential energy by
$$V=-q\int_C \mathbf{E}\cdot\mathbf{dl}$$ The quantity $$\phi (\mathbf{r})\equiv \frac{V}{q}=-\int_C \mathbf{E}\cdot\mathbf{dl}$$ is the electric potential and the path $C$ goes from a point with zero potential to $\mathbf{r}$. In the special case of conservative electric fields ($\nabla\times\vec{E}=0$, static fields, the integral above does not depend on the path $C$) you can write the electric field as
$$\mathbf{E}(\mathbf{r})=-\nabla\phi(\mathbf{r})$$
Note that in this derivation nothing has been said about the chosen coordinate system or any directions. That's why this equation is general, independent of the choice of coordinate system. This is true for all fundamental laws in physics. Equations which depend on the chosen coordinate system you would only find if you study a concrete system which is asymmetric. However in this case these would be equations which apply to this concrete system only and not general equations.
Example point charge at origin
For a charge $q$ at the origin, the electric field is given by Coulomb's law as:
$$\mathbf{E}(\mathbf{r})=\frac{kq}{r^2}\hat{\mathbf{r}}$$
where $\hat{\mathbf{r}}=\frac{\mathbf{r}}{|\mathbf{r}|}$ is the unit vector from the origin to the point $\mathbf{r}$. The corresponding electric potential is
$$\phi(\mathbf{r})= \frac{kq}{r}$$
and only depends on the distance $r$ from the origin (not on the direction), because the system is spherically symmetric.
The electric field satisfies the symmetry:
$$\mathbf{E}(\mathbf{-r})=-\mathbf{E}(\mathbf{r}),$$
telling you that the electric field points in opposite directions and is of equal strength for points $\mathbf{r}$ that are point-symmetric to the origin.
The electric potential satisfies the symmetry:
$$\phi(-\mathbf{r})=\phi(\mathbf{r})$$
In this spherical symmetric system, the gradient simplifies to:
$$\mathbf{E}(\mathbf{r})=-\nabla\phi(\mathbf{r})=-\frac{\partial\phi}{\partial r}\hat{\mathbf{r}}$$, which as you can see is satisfied by the expressions for $\mathbf{E},\phi$ above for all $\mathbf{r}$.
Change of coordinate system
If you transform to a new coordinate system where
$$\mathbf{r}'=-\mathbf{r}$$ and consequently (handwaving physicist way of doing things, if you are a mathematician take a deep breath now... or look up Jacobian which would also work for more general coordinate transformations)
$$\partial r' = - \partial r$$
you get
$$\mathbf{E}(\mathbf{r}')=\mathbf{E}(-\mathbf{r})
=\nabla\phi(\mathbf{r})
=-\nabla'\phi(\mathbf{r}')$$
where I have used $\nabla'=-\nabla$ to denote the gradient in the new coordinate system. So I guess your problem might be because you forgot to transform the gradient to the new coordinate system, which you should. As I suggested above, for general coordinate transforms the Jacobian will enter and you could try for yourself to see whether you can for instance rotate your coordinate system or whether you can transform to cartesian coordinates and still recover the same general equation $\mathbf{E}=-\nabla\phi$ (you should!).
(B)Also if you move in the direction of (−r) electric field and potential will increase.
I don't understand this argument. Could you elaborate what you mean?
