My answer builds up from the hint that user297024 gave, which is crucial in my opinion. The "previous part of the question" that you proved $H \in \operatorname{Fix}(x)$ will be needed to finish my proof.
I will assume that the Orbit-Stabilizer Theorem can be used.
By the Orbit-Stabilizer Theorem for any element $x$ of a set $X$ acted by $H$, we have:
$$
|H| = |\operatorname{Stab}(x)|. |\operatorname{Orb}(x)|
$$
As $|H| = p$ is a prime number, that leaves either $(|Stab(x)| ; |Orb(x)|)$ to be $(1;p)$ or $(p;1)$. In other words, either $\operatorname{Stab}(x) = H$ or $\operatorname{Stab}(x) = {id}$.
We will prove this by contradiction. Suppose $H$ does not fix the whole set $X$: $\exists x_{1} \neq x_{2} \in X$ such that $\operatorname{Stab}(x_{1}) = H$ but $\operatorname{Stab}(x_{2}) = id$.
This implies that $\operatorname{Orb}(x_{2}) = X$ - every element in X can be "matched" from $x_{2}$ by some element of $H$, including $x_{1} \in X$. (Note that this statement is only valid if $|X|=|H|=p$.)
Therefore, $\exists h \in H$ such that:
$$
h * (x_{2}) = x_{1}
$$
As $h \in H$, then also $h^{-1} \in H.$ Consider the following:
$$
(h^{-1}h) * x_{2} = id * x_{2} = x_{2}
$$
by the definition of a group action. But
$$
h^{-1} * (h * x_{2}) = h^{-1} * x_{1} = x_{1}
$$
by the way we construct $x_{1}$.
This leads to a violation of the definition of a group action. Therefore either $H$ fixes the whole set $X$ or $H$ does not fix anything in $X$. Now in the context of the problem, $X = G/H$ with exactly
$$
|X|=|G/H|= \frac{|G|}{|H|}=\frac{p^{2}}{p}=p
$$
by Lagrange's Theorem, and we know for a fact that $H$ is a coset of $G/H$. So using the result $H \in \operatorname{Fix}(H)$, that means every other cosets must be fixed by $H$. This completes the proof $\operatorname{Fix}(H) = G/H.$
