Automorphism covering space and universal cover

I have again a new question about the automorphism of covering space and the universal cover of a topological space $$B$$. Actually, let $$p : X \rightarrow B$$ the universal cover of $$B$$. I take a normal subgroup $$H$$ of $$Aut(p)$$, and we consider : $$p_H : X/H \rightarrow B$$. I would like to prove :

$$Aut(p_H) \cong Aut(p)/H$$

I know that as $$p_H$$ is a Galois covering space, then : $$Aut(p_H) \cong \pi_1(B,b)/(\pi_H)_*(\pi_1(X/H, x)$$, where $$x \in p_H^{-1}(b)$$.

We know that $$Aut(p) \cong \pi_1(B,b)$$, let say by a isomorphism $$f$$, and then we would like to say that $$f(H) = (p_H)_*(\pi_1(X/H, x))$$, and that's where I'm stuck.

Someone could help me, please ?

Thank you !

You don't have to use the isomorphism $$Aut(p)\simeq\pi_1(B,b)$$ here. When you see the result, you could think about creating a morphism $$Aut(p)\to Aut(p_H)$$ with kernel $$H$$. Lets try to do so. Let $$f\in Aut(p)$$, we want to find/create a function $$f^\prime\in Aut(p_H)$$. We can proceed as follows:
There is a unique $$g\in Aut(p_H)$$ such that the following diagram is commutative $$\require{AMScd} \begin{CD} X @>{f}>> X\\ @V{\pi}VV @VV{\pi}V\\ X/H @>g>> X/H \end{CD}$$ where $$\pi:X\to X/H$$ is the natural projection.
Let $$f^\prime$$ be this unique element. $$f^\prime$$ is just the factorization of $$\pi\circ f$$ by $$\pi$$ (but it is not obvious at first that such a factorization exists, this comes from the hypothesis that $$H$$ is normal in $$Aut(p)$$).
You can now check that the mapping $$\phi:f\mapsto f^\prime$$ is a morphism from $$Aut(p)$$ to $$Aut(p_H)$$, by using uniqueness in the preceding proposition. If we prove that $$H=Ker(\phi)$$ we are done.
First by definition of $$H$$, every element $$f\in H$$ verifies $$\pi\circ f=\pi$$ which can be rewritten as $$\pi\circ f=id_{X/H}\circ\pi$$ which means $$\phi(f)=id_{X/H}$$, so $$H\subset Ker(\phi)$$. Now if $$f\in Ker(\phi)$$, let $$x\in X$$. Because $$\pi(x)=\pi(f(x))$$ we have $$x\sim_H f(x)$$ so there exists $$h\in H$$ such that $$h(x)=f(x)$$. Hence $$f$$ and $$h$$ are two covering transformations which coincide at one point, so $$f=h\in H$$ which means $$ker(\phi)\subset H$$. We just proved $$H=ker(\phi)$$.