[proofplan]
We prove both isomorphisms by constructing explicit maps and verifying bijectivity. For the endomorphism ring: by the [Structure of Division Points](/theorems/???), $F(n) \cong \mathcal{O}_K/\pi^n\mathcal{O}_K$ as $\mathcal{O}_K$-modules, so $\operatorname{End}_{\mathcal{O}_K}(F(n)) \cong \operatorname{End}_{\mathcal{O}_K}(\mathcal{O}_K/\pi^n\mathcal{O}_K) \cong \mathcal{O}_K/\pi^n\mathcal{O}_K$ (since the endomorphism ring of a free rank-$1$ module over a commutative ring is the ring itself). For the automorphism group: the units of $\mathcal{O}_K/\pi^n\mathcal{O}_K$ are the images of $\mathcal{O}_K^\times$, and $\ker(\mathcal{O}_K^\times \to (\mathcal{O}_K/\pi^n\mathcal{O}_K)^\times) = U_K^{(n)} = 1 + \pi^n\mathcal{O}_K$, giving the isomorphism $U_K/U_K^{(n)} \cong \operatorname{Aut}_{\mathcal{O}_K}(F(n))$.
[/proofplan]
[step:Identify $\operatorname{End}_{\mathcal{O}_K}(F(n))$ with $\mathcal{O}_K/\pi^n\mathcal{O}_K$]
By the [Structure of Division Points](/theorems/???), there exists an $\mathcal{O}_K$-module isomorphism $\bar{\varphi} \colon \mathcal{O}_K/\pi^n\mathcal{O}_K \xrightarrow{\sim} F(n)$ sending $a + \pi^n\mathcal{O}_K \mapsto [a]_F(\lambda_n)$ for a fixed generator $\lambda_n \in F(n) \setminus F(n-1)$.
Define the map
\begin{align*}
\Psi \colon \mathcal{O}_K/\pi^n\mathcal{O}_K &\to \operatorname{End}_{\mathcal{O}_K}(F(n)) \\
a + \pi^n\mathcal{O}_K &\mapsto \left(x \mapsto [a]_F(x)\right).
\end{align*}
This is well-defined: if $a \equiv a' \pmod{\pi^n}$, then $a - a' \in \pi^n\mathcal{O}_K$, so $[a - a']_F(x) = [\pi^n]_F([c]_F(x)) = 0$ for all $x \in F(n)$ (where $a - a' = c\pi^n$), using $[\pi^n]_F(y) = 0$ for all $y \in F(n)$. Therefore $[a]_F(x) = [a']_F(x)$ for all $x \in F(n)$.
The map $\Psi(a)$ is an $\mathcal{O}_K$-module endomorphism of $F(n)$: for $x, y \in F(n)$ and $b \in \mathcal{O}_K$,
\begin{align*}
[a]_F(x +_F y) &= [a]_F(F(x, y)) = F([a]_F(x), [a]_F(y)) = [a]_F(x) +_F [a]_F(y), \\
[a]_F([b]_F(x)) &= [ab]_F(x) = [ba]_F(x) = [b]_F([a]_F(x)),
\end{align*}
where the first identity uses the endomorphism property of $[a]_F$ and the second uses $[ab]_F = [a]_F \circ [b]_F$ and commutativity of $\mathcal{O}_K$.
$\Psi$ is a ring homomorphism: $\Psi(1) = \operatorname{id}_{F(n)}$ since $[1]_F = \operatorname{id}$, and $\Psi(a)\Psi(b) = \Psi(ab)$ since $[a]_F \circ [b]_F = [ab]_F$.
**Injectivity.** Suppose $\Psi(a) = 0$, i.e., $[a]_F(x) = 0$ for all $x \in F(n)$. In particular, $[a]_F(\lambda_n) = 0$. By the kernel computation in the proof of the [Structure of Division Points](/theorems/???), this implies $a \in \pi^n\mathcal{O}_K$, so $a + \pi^n\mathcal{O}_K = 0$.
**Surjectivity.** Let $\phi \in \operatorname{End}_{\mathcal{O}_K}(F(n))$. Since $\lambda_n$ generates $F(n)$ as an $\mathcal{O}_K$-module (via $\bar{\varphi}$), the endomorphism $\phi$ is determined by $\phi(\lambda_n) \in F(n)$. By surjectivity of $\bar{\varphi}$, there exists $a \in \mathcal{O}_K$ with $\phi(\lambda_n) = [a]_F(\lambda_n)$. For any $x \in F(n)$, write $x = [b]_F(\lambda_n)$ for some $b \in \mathcal{O}_K$ (again by surjectivity of $\bar{\varphi}$). Then
\begin{align*}
\phi(x) = \phi([b]_F(\lambda_n)) = [b]_F(\phi(\lambda_n)) = [b]_F([a]_F(\lambda_n)) = [ba]_F(\lambda_n) = [a]_F([b]_F(\lambda_n)) = [a]_F(x),
\end{align*}
where we used the $\mathcal{O}_K$-linearity of $\phi$ in the second equality. Therefore $\phi = \Psi(a)$.
[guided]
The argument is a standard algebraic fact: the endomorphism ring of a free rank-$1$ module over a commutative ring $R$ is $R$ itself, acting by scalar multiplication. Here $R = \mathcal{O}_K/\pi^n\mathcal{O}_K$ and $F(n) \cong R$ as $R$-modules.
Why does it work concretely? Any $\mathcal{O}_K$-endomorphism $\phi$ of $F(n)$ is determined by $\phi(\lambda_n)$ because $\lambda_n$ generates $F(n)$: every element $x \in F(n)$ has the form $x = [b]_F(\lambda_n)$ for some $b \in \mathcal{O}_K$, and $\mathcal{O}_K$-linearity forces $\phi(x) = [b]_F(\phi(\lambda_n))$. If $\phi(\lambda_n) = [a]_F(\lambda_n)$, then $\phi(x) = [b]_F([a]_F(\lambda_n)) = [ab]_F(\lambda_n) = [a]_F([b]_F(\lambda_n)) = [a]_F(x)$, so $\phi = [a]_F|_{F(n)}$.
For injectivity, if $[a]_F$ acts as zero on all of $F(n)$, it kills $\lambda_n$ in particular. The kernel computation from the [Structure of Division Points](/theorems/???) proof showed that $[a]_F(\lambda_n) = 0$ forces $v_K(a) \geq n$, i.e., $a \in \pi^n\mathcal{O}_K$.
[/guided]
[/step]
[step:Identify $\operatorname{Aut}_{\mathcal{O}_K}(F(n))$ with $U_K/U_K^{(n)}$]
An endomorphism $\Psi(a)$ is an automorphism if and only if it is bijective. Since $F(n)$ is finite, bijectivity is equivalent to injectivity, which is equivalent to: $[a]_F(x) = 0$ for $x \in F(n)$ implies $x = 0$.
**Claim:** $\Psi(a)$ is an automorphism if and only if $a \in \mathcal{O}_K^\times$ (i.e., $v_K(a) = 0$).
*Forward direction.* Suppose $v_K(a) \geq 1$, i.e., $a = \pi c$ for some $c \in \mathcal{O}_K$. Then for any $\lambda_1 \in F(1) \setminus \{0\}$, we have $[a]_F(\lambda_1) = [\pi c]_F(\lambda_1) = [\pi]_F([c]_F(\lambda_1))$. Since $[c]_F(\lambda_1) \in F(1)$ (as $F(1)$ is an $\mathcal{O}_K$-submodule of $F(n)$) and $[\pi]_F$ kills $F(1)$, we get $[a]_F(\lambda_1) = 0$ with $\lambda_1 \neq 0$. So $\Psi(a)$ is not injective, hence not an automorphism.
*Backward direction.* Suppose $a \in \mathcal{O}_K^\times$. The endomorphism $[a]_F$ is an automorphism of the entire formal module $\bar{\mathfrak{m}}_F$ with inverse $[a^{-1}]_F$ (since $[a]_F \circ [a^{-1}]_F = [aa^{-1}]_F = [1]_F = \operatorname{id}$). In particular, $[a]_F$ restricted to $F(n)$ maps $F(n)$ bijectively onto $[a]_F(F(n))$. We verify $[a]_F(F(n)) = F(n)$: for $x \in F(n)$, $[\pi^n]_F([a]_F(x)) = [a\pi^n]_F(x) = [a]_F([\pi^n]_F(x)) = [a]_F(0) = 0$, so $[a]_F(x) \in F(n)$. Surjectivity onto $F(n)$ follows from the same argument applied to $a^{-1}$: $[a^{-1}]_F(F(n)) \subset F(n)$, so $F(n) = [a]_F([a^{-1}]_F(F(n))) \subset [a]_F(F(n))$.
Under the isomorphism $\Psi$, the automorphism group corresponds to the unit group:
\begin{align*}
\operatorname{Aut}_{\mathcal{O}_K}(F(n)) = \{\Psi(a) : a \in \mathcal{O}_K^\times\} \cong \mathcal{O}_K^\times / (\mathcal{O}_K^\times \cap \pi^n\mathcal{O}_K).
\end{align*}
Since $\mathcal{O}_K^\times \cap \pi^n\mathcal{O}_K = \{u \in \mathcal{O}_K^\times : u \equiv 1 \pmod{\pi^n} \text{ in the sense that } u - 1 \in \pi^n\mathcal{O}_K\}$... but we need to be more precise. The kernel of $\Psi|_{\mathcal{O}_K^\times}$ is $\{u \in \mathcal{O}_K^\times : \Psi(u) = \operatorname{id}\}$. Now $\Psi(u) = \operatorname{id}$ means $[u]_F(x) = x$ for all $x \in F(n)$, i.e., $[u - 1]_F(x) = [u]_F(x) -_F x = 0$ for all $x \in F(n)$ (using $[u]_F(x) -_F x = [u]_F(x) +_F [-1]_F(x) = [u - 1]_F(x)$ is not quite correct; instead, $\Psi(u) = \Psi(1)$ means $\Psi(u - 1) = 0$, i.e., $u - 1 \in \pi^n\mathcal{O}_K$).
The kernel is therefore $\{u \in \mathcal{O}_K^\times : u - 1 \in \pi^n\mathcal{O}_K\} = 1 + \pi^n\mathcal{O}_K =: U_K^{(n)}$, and we obtain the isomorphism
\begin{align*}
U_K/U_K^{(n)} = \mathcal{O}_K^\times / (1 + \pi^n\mathcal{O}_K) \xrightarrow{\;\sim\;} \operatorname{Aut}_{\mathcal{O}_K}(F(n)).
\end{align*}
[guided]
The passage from endomorphisms to automorphisms uses a valuation argument. If $v_K(a) \geq 1$, then $[a]_F$ factors through $[\pi]_F$, which kills $F(1) \subset F(n)$, so $[a]_F$ has a nontrivial kernel on $F(n)$ and cannot be an automorphism. Conversely, if $a \in \mathcal{O}_K^\times$, then $[a]_F$ has a two-sided inverse $[a^{-1}]_F$, so it is an automorphism of the whole formal module and restricts to an automorphism of $F(n)$ (one must check that $[a]_F$ preserves $F(n)$, which follows from $[\pi^n]_F \circ [a]_F = [a\pi^n]_F = [a]_F \circ [\pi^n]_F$).
For the kernel of $\mathcal{O}_K^\times \to \operatorname{Aut}_{\mathcal{O}_K}(F(n))$: a unit $u$ acts as the identity on $F(n)$ if and only if $\Psi(u) = \Psi(1)$, i.e., $\Psi(u - 1) = 0$ (since $\Psi$ is a ring homomorphism). By injectivity of $\Psi$ on $\mathcal{O}_K/\pi^n\mathcal{O}_K$, this means $u - 1 \equiv 0 \pmod{\pi^n}$, i.e., $u \in 1 + \pi^n\mathcal{O}_K = U_K^{(n)}$. The first isomorphism theorem gives $\mathcal{O}_K^\times / U_K^{(n)} \cong \operatorname{Aut}_{\mathcal{O}_K}(F(n))$.
Why is this group $U_K/U_K^{(n)}$ and not $\mathcal{O}_K^\times / U_K^{(n)}$? The notation $U_K$ denotes the group of units $\mathcal{O}_K^\times = \{x \in K : |x| = 1\}$, which is the same as $\mathcal{O}_K^\times$. So $U_K/U_K^{(n)} = \mathcal{O}_K^\times / (1 + \pi^n\mathcal{O}_K)$, and the two notations agree.
[/guided]
[/step]
