Question. Suppose $G$ acts on $X$ (via $\Psi$) and on $Y$ (via $\Phi$), and let $f : X\to Y$ be an equivariant map ($f(g\cdot x) = g\cdot f(x)$ for all $x$ in $X$ and $g$ in $G$). Is there a formula relating the cycle indices $Z_\Psi$ and $Z_\Phi$? For example, is there some substitution which transforms one of these polynomials into the other?
Background. Though the cycle index (polynomial) is usually defined for permutation groups, there is no loss to extend it to group actions: if $G \curvearrowright X$ then there is a group homomorphism $\Psi : G \to \mathfrak{S}_X$ and each $\Psi(g)$ partitions $X$ into cycles. The cycle type $c(g)$ of $\Psi(g)$ is then an $n$-tuple (here, $n = |X|$) of non-negative integers $c_l$ which count the number of cycles of length $l$ that $\Psi(g)$ has (as a permutation of $X$). We encode all this data in the cycle index: $$Z_\Psi(z_1, \dotsc, z_n) = \frac{1}{|G|}\sum_{g \in G} z^{c(g)}$$ where we use the shorthand $z^{(c_1, ..., c_l)} = z_1^{c_1} ... z_n^{c_n}$. Note that we do not have to assume that the action is faithful; if $H = \ker\Psi$ and $\overline{\Psi}:G/H\to \mathfrak{S}_X$ be the induced action, then $Z_{\overline{\Psi}} = Z_\Psi$ because $c(\bar{g}) = c(g)$ and $|G/H| = |G|/|H|$.
There is probably no answer. Working with a few examples led me to realize that the cycle index polynomial is already "reduced" in the sense that it's not the most general invariant associated with an action.
When computing $Z_{G \curvearrowright X}$, the standard simplifying observation is to notice that, since the monomial $z^{c(g)}$ depends only on the cycle type of the permutation of $X$ induced by $g$, it suffices to sum over all possible cycle types, or, what is equivalent, over all possible partitions of $X$.
The "right" way to compute $Z_{G \curvearrowright X}$ is to sum over all conjugacy classes of $G$. Conjugate elements always have the same cycle type because their induced permutations of $X$ are conjugate, and so they have identical dynamics. But the partition of $G$ arising from the equivalence relation "$\Psi(g)$ and $\Psi(g')$ have the same cycle type" is usually coarser than the conjugacy relation, and there are situations where a cycle type class will split along an equivariant map! I did not expect this—I tacitly assumed that cycle type would be preserved—which is why I believed a formula relating $Z_{G \curvearrowright X}$ and $Z_{G \curvearrowright Y}$ might exist.
Here is an example. Let $n \ge 4$ be even, $G = D_n = \langle r, s \mid r^n, s^2, (sr)^2\rangle$, and $X = C_n = \langle c \mid c^n\rangle$. Let $G$ act on $X$ by $r\cdot x = cx$ and $s\cdot x = x^{-1}$. This is the standard faithful action of $D_n$ on an $n$-gon; I incorporated the group structure of $C_n$ to simplify my calculations.
The conjugacy classes of $D_n$ are $\{1\}$, $\{r^k, r^{-k}\}$ for $1 \le k < n/2$, $\{r^{n/2}\}$, $s\langle r\rangle$, and $sr\langle r\rangle$. In the above action of $D_n$ on $C_n$, the group elements $r^{n/2}$ and $sr$ have the same cycle type (namely, $z_2^{n/2}$) so their cycle type class comprises the two conjugacy classes $\{r^{n/2}\}$ and $\{sr, sr^3, sr^5, \dotsc\}$.
Now let $d$ be a positive divisor of $n$ and consider the subgroup $H = \langle c^d \rangle$ of $C_n$. If we are thinking of the elements of $C_n$ as vertices of an $n$-gon, then $H$ is every $d$th vertex starting from $1$. Let $Y = C_n/\langle c^d\rangle$. The obvious action of $D_n$ on $Y$ (i.e. $g\cdot(xH) = (g\cdot x)H$) actually works, and the natural map $X \to Y$ sending $x$ to $xH$ is equivariant!
Under this new action, the elements $r^{n/2}$ and $sr$ behave differently. The cycle type of $r^{n/2}$ is $z_1^d$ or $z_2^{d/2}$ according as $n/d$ is even or odd. The cycle type of $sr$ is $z_2^{d/2}$ or $z_1z_2^{(d-1)/2}$ according as $d$ is even or odd (the sole fixed point being $c^{(d-1)/2}\langle c^d\rangle$ in the latter case). So, for example, with $n=4$ and $d=2$, thinking of a square and its two diagonals, the $z_2^2$-class $\{r^2, sr, sr^3\}$ splits, with $r^2$ joining $1$, $s$, and $sr^2$ in the $z_1^2$-class while $sr$ and $sr^3$ join $r$ and $r^3$ in the $z_2$-class!