We already have an accepted answer, but since i had already started an answer and was at the half of the route beyond getting the essence of the structure, i completed it now, since it may be useful in similar contexts.
On the mathematical side the situation is as follows, recalled for the convenience of the reader from the literature.
Notations are as in the already cited paper:
Families of Elliptic Curves over Cubic Number Fields with Prescribed Torsion Subgroups, Daeyeol Jeon, Chang Heon Kim, And Yoonjin Lee
A further reference that should not be omitted is:
Markus Reichert, Explicit Determination of Nontrivial Torsion Structures of Elliptic Curves Over Quadratic Number Fields
In order to produce an example of a curve $E$ with torsion $\Bbb Z/18$, the Ansatz is to work with the Tate normal form, consider curves $E=E(b,c)$ parametrized by two
algebraic numbers $b,c$ from a cubic number field $K$,
$$
E = E(b, c)\ :\qquad y^2 + (1 − c)xy − by = x^3 − bx^2\ ,
$$
and arrange that the point
$P = (0, 0)$ has order $18$.
For this, pick two parameters $(U,V)$ satisfying the equation for $X_1(18)$:
$$
\begin{aligned}
X_1(18) \ :\qquad
g_{18}(U,V) &= 0\ ,\qquad\text{ where}
\\
g_{18}(U,V)
&:=(U-1)^2 V^2 - (U^3 - U + 1)V + U^2(U - 1) \\
&= U^3(1-V) + U^2 (V^2 -1) + U(V-2V^2) + (V^2-V)\\
&\sim_{\Bbb Q(V)^\times}
U^3 - U^2(V + 1) + \frac{2V^2-V}{V-1} -V\ .
\ .
\end{aligned}
$$
Seen as a polynomial in $U$, it has degree $3$.
We set $V=t$ to be a "suitable" rational number,
and the polynomial $g_{18}(U,t)$ defines a cubic field $K=\Bbb Q(\alpha_t)$
generated by some $\alpha_t$. Let me plot the connection to $X_1(18)$ explicitly:
$$
g_{18}(\alpha_t,t)=0\ .
$$
Then the formulas for $b,c$ are given by one and the same
rational function in $(U,V)=(\alpha_t,t)$. They are:
$$
\begin{aligned}
b(U,V) &=
-\frac
{V(U - V)(U^2 + V)(U^2 -UV + V)}
{(U^2 -V^2+V)(U^2 + UV -V^2 + V)^2}
\ ,
\\
c(U,V) &=
-\frac
{V(U - V)(U^2 -UV + V)}
{(U^2 -V^2+V)(U^2 + UV -V^2 + V)}
\ .
\end{aligned}
$$
Warming up. We proceed as follows in the given context from above.
We fix some $t$. To have a concrete example, $t$ may be specialized to $t=t_1=1/5$, as the OP does it also. Let $t'$ be its cousin,
$$
t'=t_3=\frac 1{ 1-t }\ .
$$
We build the corresponding field $K=\Bbb Q(\alpha)$, where $\alpha =\alpha_t$ is a suitable root of the polynomial $g_{18}(U,t)$, seen as a polynomial in $U$. Let $K'=\Bbb Q(\alpha')$ be the cousin field, where $\alpha'$ is a specific root for $g_{18}(U, t')$.
Question: Are $K$ and $K'$ isomorphic (for some good choice of $\alpha'$)?
Answer: Yes, they are, take $\displaystyle \alpha'= 1-\frac 1\alpha$.
To illustrate the situation, we consider first the sample case $t=1/5$.
Sage gives this information as follows:
def g18(U, V):
return (U^3*V - U^2*V^2 - U^3 + 2*U*V^2 + U^2 - U*V - V^2 + V)
R.<U> = PolynomialRing(QQ)
t1 = 1/5
a1 = g18(U, t1).roots(ring=QQbar, multiplicities=False)[0]
t3 = 1/(1 - t1)
a3 = 1 - 1/a1
print(f'g18(a3, t3) = {g18(a3, t3)}')
The sage interpreter gives after a copy+paste of the above code, together with one more line to be sure we get a clean zero:
g18(a3, t3) = 0.?e-17
sage: g18(a3, t3).minpoly()
x
Because of the rôle of $(\alpha,t)$ as a special value for $(U,V)$,
i will use below rather $(u,v)$ pairs instead. Now the whole context can be explained structurally as follows.
Proposition:
Let $F$ be a field (of characteristic $\ne 2,3$).
For two parameters $b,c\in F$, $b\ne 0$, let $E_T(b,c)$ be the elliptic curve in Tate normal form
$$
E_T(b, c)\ :\qquad y^2 +(1-c)xy -by = x^3 bx^2\ ,
$$
so that $P=(0,0)$ is a rational point on it.
For suitable ($\Delta(A,B)\ne 0$) parameters $A,B\in F$ let consider also the elliptic curve in short Weierstrass form
$$
E_W(A,B)\ :\qquad y^2 = x^3 + Ax+B\ .
$$
Fix $u,v$ în $F$, $u\ne 1$, so that the pair
$(u,v)$ corresponds to a point on the moduli space $X_1(18)$ parametrized
as mentioned above, i.e. it satisfies
$$
g_{18}(u,v)=0\ ,\qquad\text{ where }\\
g_{18}(U,V)=
(U-1)^2 V^2 -(U^3 -U + 1)V + U^2(U - 1)\ .
$$
Then the pair $(u',v')$ with components
$$
\begin{aligned}
u' &= \frac 1{1-u}\ ,\\
v' &= 1-\frac 1v\ ,
\end{aligned}
$$
is also defining a pointin the moduli space $X_1(18)$,
i.e. $g_{18}(u',v')=0$. Let $\underline A$, $\underline B$
be the rational functions given by
$$
\begin{aligned}
\underline A(b,c) &= -\frac 1{48}\Big(\ ((c-1)^2 - 4b)^2 - 24b(c - 1)\ \Big)\ ,\\
\underline B(b,c) &= \frac 1{864}\Big(\ ((c-1)^2 - 4b)^3 - 36b(c-1)^3 + 72b^2(2c + 1)\ \Big)\ .
\end{aligned}
$$
Consider with a slight abuse of notation $b,c\in F$ and $b',c'\in F$, then $A,B\in F$ and $A',B'\in F$ as follows
$$
\begin{aligned}
b &= b(u,v)\ ,\qquad &b' &= b(u',v')\ ,\\
c &= c(u,v)\ ,\qquad &c' &= c(u',v')\ ,\\[2mm]
A &= \underline A(b,c)\ ,\qquad &A' &=\underline A(b',c')\ ,\\
B &= \underline B(b,c)\ ,\qquad &B' &=\underline B(b',c')\ ,\\[2mm]
&\qquad\text{ and consider the elliptic curves}\\[2mm]
E_T &= E(b, c)\ , \qquad &E'_T &= E_T(b', c')\\
E_W &= E(A, B)\ , \qquad &E'_W &= E_W(A', B')\ .
\end{aligned}
$$
Then$$
\frac {A'}A = U^{12}\ ,\qquad
\frac {B'}B = U^{18}\ ,
$$
so the elliptic curves $E_W$ and $E_W'$ are canonically isomorphic via a map $\Phi$, as shown in the diagram below. The functions
$\underline A$, $\underline B$ were chosen to make
$E_T(b,c)$ isomorphic $E_W(A,B)$. Then the following diagram is commutative:
$\require{AMScd}$
$$
\begin{CD}
E_T @>{\cong}>> E_W\\
@A{\cong} AA @A\cong A\Phi A\\
E'_T @>>\cong> E'_W
\end{CD}
$$
So we can compare the rational points $P=(0,0)\in E_T(F)$ and $P'=(0,0)\in E'_T(F)$ in one or any of the common worlds, e.g. in $E_W(F)$, and then $11P$ and $P'$
(or equivalently $P=5\cdot 11 P$ and $5P'$) correspond to one and the same torsion point of order (dividing) $18$. In a diagram:
$\require{AMScd}$
$$
\begin{CD}
P_T @>{\cong}>> P_W=5\Phi(P'_W)=\Phi(5P'_W)\\
@. @A\cong A\Phi A\\
5P'_T @>>\cong> 5P'_W
\end{CD}
$$
Proof by computer.
$\square$
Code for the proof. First let us define the needed functions, and needed objects.
def bmap(U, V):
return -(U^2 - U*V + V) * (U^2 + V) * (U - V) * V / (U^2 + U*V - V^2 + V)^2 / (U^2 - V^2 + V)
def cmap(U, V):
return -(U^2 - U*V + V) * (U - V) * V / (U^2 + U*V - V^2 + V) / (U^2 - V^2 + V)
def Amap(b, c):
return -1/48 * ( ((c-1)^2 - 4*b)^2 - 24*b*(c - 1) )
def Bmap(b, c):
return 1/864 * ( ((c-1)^2 - 4*b)^3 - 36*b*(c-1)^3 + 72*b^2*(2*c + 1) )
def f(U, V):
return (U^3*V - U^2*V^2 - U^3 + 2*U*V^2 + U^2 - U*V - V^2 + V)
R.<U,V> = PolynomialRing(QQ)
Q = R.quotient( f(U, V) )
FR = R.fraction_field()
FQ = Q.fraction_field()
u1, v1 = FQ(U), FQ(V)
u2, v2 = 1/(1 - u1), 1 - 1/v1
Now we can check:
print(f'Is f(u2, v2) zero? {bool( f(u2, v2) == 0 )}' )
b , c = bmap(U , V ), cmap(U , V )
b1, c1 = bmap(u1, v1), cmap(u1, v1)
b2, c2 = bmap(u2, v2), cmap(u2, v2)
A , B = Amap(b , c ), Bmap(b , c )
A1, B1 = Amap(b1, c1), Bmap(b1, c1)
A2, B2 = Amap(b2, c2), Bmap(b2, c2)
print(f'Is A2/A1 = u1^12? {bool( A2/A1 == u1^12 )}')
print(f'Is B2/B1 = u1^18? {bool( B2/B1 == u1^18 )}')
ET = EllipticCurve(FR, [1 - c, -b, -b, 0, 0])
EW = EllipticCurve(FR, [A, B])
phi = ET.isomorphism_to( EW )
PT = ET.point( (0, 0, 1) )
PW = phi( PT )
five_PW = 5*PW
x_PW , y_PW = PW.xy()
x_five_PW, y_five_PW = five_PW.xy()
x_PW.subs({U: u1, V:v1}) == x_five_PW.subs({U: u2, V: v2}) / u1^6
y_PW.subs({U: u1, V:v1}) == y_five_PW.subs({U: u2, V: v2}) / u1^9
This gives the needed confirmations:
Is f(u2, v2) zero? True
Is A2/A1 = u1^12? True
Is B2/B1 = u1^18? True
True
True
The last two True values confirm that the coordinates of $P=(0,0)=E_T(F)$ and $5P'$ where $P'=(0,0)\in E_T'$ are the same, when transported to $E_W(F)$.
Note: Unfortunately, sage cannot build the needed curves over FQ.
Best Answer
If I understand correctly: you are starting with $a = 10656\sqrt{-23} +13600$, and you want to choose $f$ in the same number field such that $af^2$ is "as nice as possible", ideally lying in $\mathbf{Z}$; and you have observed that you can get it down to $a' = -28\sqrt{-23} + 605$.
Let's consider the prime factorisation of $a'$. It has the shape $P_1^2 P_2 P_3$ where $P_i$ are primes, and the prime $P_2 = \langle 139, (33 + \sqrt{-23})/2\rangle$ is not equal to its Galois conjugate $P_2^\sigma$. So you can see straight away that no element of the form $a' f^2$ can be in $\mathbf{Z}$, because it will always have even valuation at $P_2^\sigma$ and odd valuation at $P_2$, while any element of $\mathbf{Z}$ has to have equal valuations at $P_2$ and $P_2^\sigma$.
So the answer to question 1 is "no"; and, insofar as I can understand what question 3 is asking, the answer is "it doesn't matter" (the class number is not the obstruction here). For question 2, I suspect the answer is also "no", but it sounds like a mighty tedious check.