Take the $53^{rd}$ cyclotomic ring of integers and the homomorphism $\sigma: \zeta\to\zeta^2$ with the primitive root $2$ modulo $53$. Kummer's class number formula gives the prime $4889$. Because every ideal is invariant under the homomorphism $\sigma^{52}\equiv id$ and because the class group must be cyclic, the homomorphism $\sigma$ can only send every class $C$ to the class $C^\xi$ where $\xi$ must be a (not necessarily primitive) $52^{nd}$ root modulo $4889$ so that $\xi^{52}\equiv 1 \bmod{4889}$. There are $52$ different prime ideals $\sigma^k\langle 107, \psi_{107}(\zeta)\rangle = \langle 107, \sigma^k\psi_{107}(\zeta)\rangle$ lying over the prime $ 107 $. We only note here that the second generator $\psi_{107}(\zeta)$ of the prime ideal was already found by Kummer. A more detailed analysis of this ring then gives the class equivalence $\sigma\langle 107, \psi_{107}(\zeta)\rangle = \langle 107, \sigma\psi_{107}(\zeta)\rangle \sim \langle 107, \psi_{107}(\zeta)\rangle^{3637}$. The 'zip code' $3637$ is of no further interest. However it is a primitive $52^{nd}$ root modulo $4889$ and we have a clear-cut relationship between the cyclic class group of this ring and the cyclic Galois group $\operatorname{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) = \{\operatorname{id} = \sigma^0, \sigma^1, \dots, \sigma^{51}\}$.
Now take the $41^{st}$ cyclotomic ring of integer, the homomorphism $\sigma: \zeta \to \zeta^6$ with the primitive root $6$ modulo $41$ and the cyclotomic integer $g_1(\zeta) = \zeta^6 - \zeta^8 - \zeta^{20}$. The ideal $\langle g_1(\zeta)\rangle$ factorizes to $\langle g_1(\zeta)\rangle = \langle 83, \psi_{83}(\zeta)\rangle^2\cdot\langle 83, σ^8\psi_{83}(\zeta)\rangle$. The cyclotomic integer $g_2(\zeta) = - \zeta^4 + \zeta^{10} + \zeta^{21} + \zeta^{28} + \zeta^{39}$ factorizes to $\langle g_2(\zeta)\rangle = \langle 83,
\sigma^8\psi_{83}(\zeta)\rangle\cdot\langle 83, \sigma^{28}\psi_{83}(\zeta)\rangle$ so that we get the class equivalence $\langle 83, \psi_{83}(\zeta)\rangle^2 \sim \langle 83, \sigma^{28}\psi_{83}(\zeta)\rangle$. Squaring this equivalence gives
$$\langle 83, \psi_{83}(\zeta)\rangle^4 \sim \langle 83, \sigma^{28}\psi_{83}(\zeta)\rangle^2 \sim \sigma^{28}[\langle 83, \psi_{83}(\zeta)\rangle^2] \sim \sigma^{28}\langle 83, \sigma^{28}\psi_{83}(\zeta)\rangle = \langle 83, \sigma^{2·28}\psi_{83}(\zeta)\rangle.$$
Consecutive squaring gives $\langle 83, \psi_{83}(\zeta)\rangle^{2^k} \sim \langle 83, \sigma^{28k}\psi_{83}(\zeta)\rangle$ and we get the general relationship $\langle 83, \psi_{83}(\zeta)\rangle^n \sim \langle 83, \sigma^{4k}\psi_{83}(\zeta)\rangle$ for an integer $n$ because $\operatorname{gcd}(28, 40) = 4$. For $k=10$ we get $\langle 83, \psi_{83}(\zeta)\rangle^{1024} \sim \langle 83, \psi_{83}(\zeta)\rangle$ or the ideal $\langle 83, \psi_{83}(\rangle)\rangle^{1023}$ is principal. Kummer's class number formula gives $11^2$ so that the class order of the ideal $\langle 83, \psi_{83}(\zeta)\rangle$ is $11$ with $1023 = 3\cdot 11\cdot 31$. Then we have 4 class subgroups $\langle 83, \sigma^{m+4k}\psi_{83}(\zeta)\rangle, m \in \{0,1,2,3\}$. But the class number is $11^2$ so that the combination of the classes of 2 subgroups gives the classes of the remaining 2 class subgroups. For example the cyclotomic integer $s_2(\zeta) = \zeta^{28} + \zeta^{36} - \zeta^{39}$ factorizes to $\langle s_2(\zeta)\rangle = \langle 83, \sigma^{14}\psi_{83}(\zeta)\rangle\cdot\langle 83, \sigma^{24}\psi_{83}(\zeta)\rangle\cdot\langle 83, \sigma^{33}\psi_{83}(\zeta)\rangle$. Therefore the homomorphism $\sigma$ sends the class of one class subgroup to the next class subgroup. And we only have a clear relationship between the homomorphism $\sigma^{4k}$ of the Galois group of this ring and the classes because the homomorphism $\sigma^4$ does not send the classes of one class subgroup to the next. However the class group $\mathbb{Z}/11\mathbb{Z} \oplus \mathbb{Z}/11\mathbb{Z}$ has been confirmed.
The matter gets worse in the $163^{rd}$ cyclotomic ring of integers. We take the homomorphism $\sigma: \zeta \to \zeta^2$ with the primitive root $2$ modulo $163$. I could only determine the class group of the prime ideals that factorize to maximal $ 27 $ conjugate prime ideals. The analysis gives a class subgroup of $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$. Take the four classes $c$, $d$, $e$, $f$, each the generator of one cyclic class subgroup $\mathbb{Z}/2\mathbb{Z}$ with class order $2$. Then each ideal of order $2$ can be assigned to a class $(c^r, d^s, e^t, f^u)$. We take the letter $ I $ for the principal ideal. The ideals $\langle q, \psi_{q}(\zeta)\rangle$ with $q = 61, 199, 347$ all have six conjugates. The analysis similar to that above gives
\begin{align*}
\langle 61, \sigma^0\psi_{61}(\zeta)\rangle &\sim (c, I, I, I)\\
\langle 61, \sigma^1\psi_{61}(\zeta)\rangle &\sim (I, I, e, I)\\
\langle 61, \sigma^2\psi_{61}(\zeta)\rangle &\sim (I, d, I, I)\\
\langle 61, \sigma^3\psi_{61}(\zeta)\rangle &\sim (I, I, I, f)\\
\langle 61, \sigma^4\psi_{61}(\zeta)\rangle &\sim (c, d, I, I)\\
\langle 61, \sigma^5\psi_{61}(\zeta)\rangle &\sim (I, I, e, f)\\
\langle 199, \sigma^0\psi_{199}(\zeta)\rangle &\sim (I, d, e, I)\\
\langle 199, \sigma^1\psi_{199}(\zeta)\rangle &\sim (I, d, I, f)\\
\langle 199, \sigma^2\psi_{199}(\zeta)\rangle &\sim (c, d, I, f)\\
\langle 199, \sigma^3\psi_{199}(\zeta)\rangle &\sim (c, d, e, f)\\
\langle 199, \sigma^4\psi_{199}(\zeta)\rangle &\sim (c, I, e, f)\\
\langle 199, \sigma^5\psi_{199}(\zeta)\rangle &\sim (c, I, e, I)\\
\langle 347, \sigma^0\psi_{347}(\zeta)\rangle &\sim (c, I, I, f)\\
\langle 347, \sigma^1\psi_{347}(\zeta)\rangle &\sim (c, d, e, I)\\
\langle 347, \sigma^2\psi_{347}(\zeta)\rangle &\sim (I, d, e, f)
\end{align*}
and we have $\langle 347, \sigma^0\psi_{347}(\zeta)\rangle \sim \langle 347, \sigma^3\psi_{347}(\zeta)\rangle$. Here the action of the Galois group on the prime ideal $(61, \sigma^0\psi_{61}(\zeta))$ does not exhaust the classes of the class subgroup $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$. It does not even form a class subgroup of $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$. Hence we have lost all relationship between the class group and the Galois group.
My experience with class groups of cyclotomic rings of integer taught me that the classes have a structure on their own though in general there does not seem to be any link between the classes and other algebraic structures of these rings.
You can get my determination of the class groups of the rings above and other cyclotomic rings here.