In Schoof's 1995 paper, Counting points on elliptic curves over finite fields, page 236, Proposition 6.1(i) states:

Let $\mathbb{E}$ be an elliptic curve over $\mathbb{F}_p$. Suppose that its $j$-invariant is not supersingular and that $j\neq 0 $ or $1728$.

Then the modular polynomial $\Phi_l(j,T)$ has a zero $\tilde{\jmath} \in \mathbb{F}_{p^r}$ if and only if the kernel $C$ of the corresponding isogeny $E \mapsto E/C$ is a one-dimensional eigenspace of $\phi^r_p$ in $E[l]$, with $\phi_p$ the Frobenius endomorphism of $E$.

From the proof:

If $\Phi_l(j,\tilde{\jmath} )=0$, then there exists a cyclic subgroup $C$ of $E[l]$ such that the $j$ invariant of $E/C$ is equal to $\tilde{\jmath}\in \mathbb{F}_{p^r} $. Let $E^1$ be an elliptic curve over $\mathbb{F}_{p^r}$ with $j$ invariant equal to $\tilde{\jmath}$. Let $E/C \mapsto E^1$ be an $\bar{\mathbb{F}}_p$ isomorphism and let $f: E \mapsto E/C \mapsto E^1$ be the composite isogeny. It has kernel $C$.

I can see $f$ is defined over $\mathbb{F}_{p^r}$, so this implies existence of an isogeny $E \mapsto E^1$ over $\mathbb{F}_{p^r}$.

1. How does this imply that $C$ is an eigenspace of $\phi^r_p$?

2. How does the statement "The Frobenius endomorphism over $\mathbb{F}_{p^r}$ satisfies the same characteristic equation" imply that $C$ is an eigenspace of $\phi^r_p$?

Please help me to understand this.