Finding a necessary and sufficient conditon for the coefficient $(p,q)$ so that $fRg$ where $f(x)=x^2$ and $g(x)=x^2+px+q$ $R$ is the relation defined in $\Bbb R^{\Bbb R}$  by:

$$fRg \Leftrightarrow \exists \psi \in \Bbb R^{\Bbb R} , \psi  \text{ bijective and }\psi \circ f = g \circ \psi$$

I know that $R$ is an equivalence relation however I don't see how I can use that in any way.
I tried to find different conditions like $(p-1)²\gt 4q$ but there's  no way to prove that they are sufficient.
Further observations, too long for a comment.

We have 
$$\psi(x^2)=\psi^2(x)+p\psi(x)+q,\tag{1}$$
which gives for $k\in\{0,1\}$ that
$$\psi(k)=\psi^2(k)+p\psi(k)+q,\tag{2}$$
from where in case of
$$(p-1)^2\geq 4q,\tag3$$
the solutions
$$\psi(k)=-\frac{p-1}{2}\pm\sqrt{\frac{(p-1)^2}{4}-q}$$
follows (which doesn't mean that $\psi(0)=\psi(1)$).  Now if we assume that $\psi$is $C^1$, we have
$$\psi'(x^2)\cdot 2x=2\psi(x)\psi'(x)+p\psi'(x),$$
which gives for $x=0$ that
$$0=\psi'(0)\bigl(2\psi(0)+p\bigr),$$
that is either $\psi'(0)=0$ or $\psi(0)=-p/2$.  Let's first deal with $\psi(0)=-p/2$ and plug it in $(2)$, which yields in
$$-\frac12=\pm\sqrt{\frac{(p-1)^2}{4}-q}.$$
Now that gives $4q=p^2-2p$ as Michael Burr stated above.  Investigating $\psi'(1)$ gives  the same result: either $\psi'(1)=0$ or $\psi(1)=(2-p)/2$ and $4q=p^2-2p$ again.  This leaves the cases $\psi'(k)=0$ open. If we further assume that $\psi$ is also $C^2$, the $\psi''(k)$ must be zero as well, otherwise $\psi$ would not be bijective.

Now we can give an example of $\psi$, namely the straight line passing through $(0,-p/2)$ and $(1,1-p/2)$:
$$\psi(x)=x-\frac p2.$$
Consistent with comments, we can observe that $\psi\left(0\right)$ is a fixed point of $g$. And another important observation is that $$\psi\left(t^2\right)=\psi\left(t\right)^2+p\psi\left(t\right)+q=\psi\left(-t\right)^2+p\psi\left(-t\right)+q$$ holds for any natural $t$. Since $g$ is quadratic, it must be $\psi\left(t\right)=\psi\left(-t\right)$ or $\psi\left(t\right)+\psi\left(-t\right)=-p$ where $t\neq0$, and $\psi$ is bijective according to assumption, so the latter holds. $\\$Now it's clearly that $\psi$ also maps $0$ to the extreme point of $g$, hence we obtain the general expression of $g$ :$$g=\left(x-a\right)^2+a$$, where $p=-2a,\,q=a^2+a$ and $a$ is an arbitrary natural number. $\\$Of course the sufficiency could be proved. In fact, let $f\left(x\right)=x^2$ and $g\left(x\right)=\left(x-a\right)^2+a$, $\psi\left(x\right)=x+a$ is an eligible mapping.