= Solution
The map
$$
[i](x,y)=(-x,iy)
$$
is an automorphism of the <elliptic curve> over $K$, fixes its identity point, and squares to $[-1]$. For the <invariant differential on an elliptic curve> $\omega=dx/(2y)$, it satisfies $[i]^*\omega=i\omega$. Thus $\mathbb Z[i]\subseteq\operatorname{End}_K(E)$, with the indicated embedding of $K$.
Under analytic <complex uniformization of an elliptic curve> $E(\mathbb C)\simeq\mathbb C/L$, this automorphism acts as multiplication by $i$, so $L$ is stable under $\mathbb Z[i]$. It is a rank-one torsion-free <module> over the <Gaussian integers>. Since that <ring> is a <principal ideal domain>, $L=\Omega\mathbb Z[i]$. Every analytic <endomorphism> is multiplication by an $a\in\mathbb C$ with $aL\subseteq L$, which here says exactly $a\in\mathbb Z[i]$. Therefore the geometric <endomorphism ring of an elliptic curve> is $\mathbb Z[i]$, and all its elements are generated over $\mathbb Z$ by the displayed $K$-defined automorphism. Thus $\operatorname{End}_K(E)=\mathbb Z[i]$.
If an <endomorphism> is defined over $\mathbb Q$, its pullback on the rational differential $\omega$ has scalar in $\mathbb Q$. The differential action is injective in characteristic zero, because a nonzero <isogeny of elliptic curves> is separable and has nonzero differential. Its scalar must therefore lie in $\mathbb Q\cap\mathbb Z[i]=\mathbb Z$. Conversely integer multiplication is defined over $\mathbb Q$. Hence
$$
\boxed{\operatorname{End}_{\mathbb Q}(E)=\mathbb Z,\qquad\operatorname{End}_{\mathbb Q(i)}(E)=\mathbb Z[i].}
$$
Put $\lambda=1+i$. The valuations of the differences of the three nonidentity fourth roots of unity from $1$ are
$$
v_\lambda(-1-1)=2,\qquad v_\lambda(i-1)=1,\qquad v_\lambda(-i-1)=1.
$$
For instance $2=-i\lambda^2$, and the other two differences are associates of $\lambda$. None is divisible by $\lambda^3$. The group $(\mathbb Z[i]/\lambda^3)^\times$ has $8-4=4$ elements, so the four units map bijectively to it. Every ideal prime to $\lambda$ consequently has a unique <primary Gaussian integer> generator.
The <Grössencharacter of the square-lattice elliptic curve> is explicitly
$$
\boxed{\psi_E(\mathfrak a)=\alpha,\qquad(\alpha)=\mathfrak a,\quad\alpha\equiv1\pmod{\lambda^3},\qquad\mathfrak f=\lambda^3.}
$$
For integral ideals prime to $2$, the generator is an odd <Gaussian integer>; extend multiplicatively to fractional ideals prime to $2$. Products of primary generators are primary, so this is a <Hecke character> with ideal infinity type $(1,0)$: on a principal ideal with generator $a\equiv1\pmod{\lambda^3}$, its value is $a$. Its conductor divides $\lambda^3$, and cannot divide $\lambda^2$, since $-1\equiv1\pmod{\lambda^2}$ would force $\psi_E(({-1}))=-1$, whereas the unit ideal has value $1$. Thus the displayed conductor is exact.
We next verify directly that this character gives the Frobenius values of this particular curve, fixing the unit ambiguity. The point
$$
P=(i,1-i)\in E(K)
$$
satisfies, by the chord-and-tangent addition law,
$$
[\lambda]P=(1,0),\qquad[\lambda]^2P=(0,0),\qquad[\lambda]^3P=O.
$$
For the first addition, $[i]P=(-i,1+i)$ and the chord has slope $-1$. Since $P$ is killed by $\lambda^3$ but not by $\lambda^2$, it generates the eight-element $\mathbb Z[i]$-module $E[\lambda^3]\simeq\mathbb Z[i]/\lambda^3$. Thus all this torsion is rational over $K$.
The model has discriminant $64$, so it has <good reduction> at every $\gamma\ne(\lambda)$. First suppose $\gamma$ lies over a split rational prime $p$. Then its residue field is $\mathbb F_p$, $p\geq5$, and $\pi=\psi_E(\gamma)$ has norm $p$. Reduction of $[\pi]$ still has degree $p$ and has zero differential, since its differential scalar $\pi$ belongs to $\gamma$. It is therefore purely inseparable of degree $p$. Factoring out the <Frobenius isogeny> gives
$$
\widetilde{[\pi]}=u\circ\operatorname{Frob}_p
$$
for an automorphism $u$ of the reduction. In characteristic at least five, the automorphisms of this model fixing $O$ are the four Gaussian units: comparing coefficients in $(x,y)\mapsto(v^2x,v^3y)$ gives $v^4=1$. Prime-to-$p$ torsion reduces injectively. Both $[\pi]$ and $\operatorname{Frob}_p$ fix the reduction of $E[\lambda^3]$, the former by the primary congruence and the latter by its rationality over the residue field. Thus $u$ fixes this entire module. The unit congruences proved above force $u=1$.
For an inert rational prime $p\equiv3\pmod4$, we have $\gamma=(p)$, $N\gamma=p^2$, and its primary generator is $\pi=-p$, because $4\mid p+1$ implies $\lambda^3\mid(-p-1)$. To determine Frobenius without excluding $p=3$, write
$$
a_p=-\sum_{x\in\mathbb F_p}\left(\frac{x^3-x}{p}\right).
$$
Replacing $x$ by $-x$ changes each nonzero quadratic-character value to its negative, because $-1$ is a nonsquare. Hence $a_p=0$. The Frobenius characteristic equation $\operatorname{Frob}_p^2-[a_p]\operatorname{Frob}_p+[p]=0$ gives $\operatorname{Frob}_{p^2}=[-p]$. This proves the asserted Frobenius value at inert primes as well, and identifies the displayed <Hecke character> with that of $E$.
Finally, the formal parameter $t=-x/y$ is defined over the local <ring> at $\gamma$. Since the reduced <endomorphism> is the $q$-power <Frobenius isogeny>, with $q=N\gamma$, its pullback sends $t$ to $(-x/y)^q=t^q$. Passing to the completion at the identity gives the equality of formal series
$$
\boxed{[\pi](t)\equiv t^{N\gamma}\pmod\gamma.}
$$
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