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I am studying $l$-isogeny graphs (volcanoes). As I understand these graphs have $j$-invariants as vertices but I am having a hard time understanding the edges. The following is not clear to me:

Suppose $E_1/\mathbb{F}_q, E_2/\mathbb{F}_q$ are elliptic curves with the same $j$-invariant $j$. If $\phi_1: E_1 \to E_1'$ is an $l$-isogeny, then there exist an $l$-isogeny $\phi_2: E_2 \to E_2'$.

I've searched for the proof but everyone is using modular polynomial. I would like to avoid modular polynomial as it seems to me that it could be done in more elementar way. Am I wrong?

This was my (maybe naive) approach: Let $\rho$ be an isomorphism over $\overline{\mathbb{F}_q}$ between $E_1$ and $E_2$. Every separable isogeny is determined by its kernel, let $G=ker \phi_1$. Since $G$ is a cyclic subgroup of $E_1[l]$ then $\rho(G)$ is a cyclic subgroup of $E_2[l]$. For $\rho(G)$ to be a kernel of $l$-isogeny it would have to be defined over $\mathbb{F}_q$ (invariant under the elements of $Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q)$). We know that $G$ is defined over $\mathbb{F}_q$. This is where my idea ends.

It would be marvelous If I could show that $E_2, E_2'$ have the same $j$-invariant along the way. I will be grateful for any tips, ideas, links, anything. Thanks a lot.

EDIT: I realized that if $j\neq 0,1728$ (so let's assume that) then $\rho$ is defined over quadratic extension $L$ of $\mathbb{F}_q$. The isomorphism can be then written as $\rho(x,y)=(c_1x+c_2, (d_1x+d_2)y)$ where $c_1,c_2,d_1,d_2 \in L$. Also it suffices to show that $\rho(G)$ is invariant under Frobenius automorphism (not every automorphism from $Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q)$). So it breaks down to prove the following: For each $P=(x,y) \in G$: $\pi(\rho(P))=\pi(c_1x+c_2, (d_1x+d_2)y)=(c_1^qx^q+c_2^q, (d_1^qx^q+d_2^q)y^q) \in \rho(G)$ where we know that $(x^q,y^q) \in G$.

Not sure what to do next, everything I tried was too messy.

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I think I figured it out. We want to show that $\rho(G)$ is invariant under Frobenius automorphism. The key fact is that every element in the finite field $L$ has a $p$th root. So there is an isogeny $\rho'$ such that $\rho\pi_p = \pi_p \rho'$ where $\pi_p$ is the map $(x,y)\mapsto (x^p,y^p)$. Comparing degrees we get that $\rho'$ is also an isomorphism. Now, the automorphism group of $E_1$ is just $\mathbb{Z}_2$ because $j\neq 0,1728$. And since $\rho^{-1}\rho'$ is an automorphism, it is necessary that $\rho^{-1}\rho'=-id$ or $id$. The rest is easy, as well as the fact that $E_2$, $E_2'$ have the same $j$-invariant (construct an isomorphism using the two isogenies and $\rho$)

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