Given $N$ with $d$ prime factors. Can the number of unique values $x^d \mod N$ calculated for $d>2$? Does the total amount decrease at some point?

Question

Given a number $N$ with $d$ unique prime factors. Can the number of unique values $v$ with $$v \equiv x^d \mod N$$ $$x\in[0,N-1]$$ $$N = \prod_{i=1}^{d} p_i$$ be calculated for $d>2$? (Q1)
Does the total amount decrease at some point? (Q2)

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For simplification we assume each prime factor $p_i > 5$.
Or for target use case each $p_i$ is big enough to avoid easy factorization.

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Solving trial:
For $d=1$ it's trivial. If we insert every value from $0$ to $N-1$ for $x$ in $x^1 \mod N$. There we always have $N$ unique values.
So $N_{x^1} = N$

For $d=2$ we have two interacting groups from $p_1$ and $p_2$ with size $p_1-1$ and $p_2-1$ with a shared prime factor of at least $2$. If we combine them we get (in most cases) a group size of
$$L = \mathrm{lcm}(\frac{p_1-1}{2}-1, \frac{p_2-1}{2}-1)$$ And a number of $L_n$ instances $$L_n = \mathrm{gcd}(\frac{p_1-1}{2}-1, \frac{p_2-1}{2}-1)$$ And some special cases for $0$, $1$, numbers with the '$\frac{p_i-1}{2}$'-th power ($\mod N$) and some special special case if the base is also $p_i^2$
With this we can calculate the total number of quadratic residue ($d=2$) $N_{x^2}$ among $\mathbb Z/N\mathbb Z$: $$N_{x^2} = L_n\cdot L + 2 + 2 (\frac{p_1-1}{2}-1)+2(\frac{p_2-1}{2}-1)+2$$ (more details in answer and question)

Q1: Is there a more general equation for $d>2$?

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Testing around:
In some test for $d \in [2,3,4,5,6]$ I computed all possible values and noticed the ratio $$R_d = \frac{N_{x^d}}{N}$$ can be $1$ for $d\in [3,5]$ but also just $0.1$. For $d=2$ it's $R_2 \approx 0.25$.
$R_4$ was always $<0.05$ in test cases. $R_6$ seems to be even smaller with some $R_6<0.001$

Q2.1: Will this ratio continue to decrease for larger (even) $d$?

Q2.2: Does the total amount of $N_{x^d}$ decrease at some $d$?
Let's assume $N$ gets 512-bit bigger for each new prime factor, will there be a $d$ (with a $d \cdot 512$-bit $N$) which has less $N_{x^d}$ than $N_{x^2}$ (with $2\cdot 512 = 1024$-bit $N$)? (Q2.3)

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Examples:
$d=2$
$N = 50471 =41\cdot 1231$ with $N_{x^2}=12936$ and $R_2 = 0.256$
$ N = 28363 = 113 \cdot 251$ with $N_{x^2}= 7182 $ and $R_2 = 0.253$

$d=3$
$N =18031=13\cdot 19\cdot 73$ with $N_{x^3}=875$ and $R_3 =0.04$
$N =11339=17\cdot 23\cdot 29$ with $N_{x^3}=11339$ and $R_3 =1.0$

$d=4$
$N =97867=7\cdot 11\cdot 31\cdot 41$ with $N_{x^4}=4224$ and $R_4=0.04$
$N =63427=7\cdot 13\cdot 17\cdot 41$ with $N_{x^4}=880$ and $R_4=0.01$

$d=5$
$N =3453307=11\cdot 13\cdot 19\cdot 31\cdot 41$ with $N_{x^5}=46683$ and $R_5=0.0135$
$N =1659931=7\cdot 13\cdot 17\cdot 29\cdot 37$ with $N_{x^5}=1659931$ and $R_5=1.0$

$d=6$
$N=28709681=7\cdot 11\cdot 13\cdot 23\cdot 29\cdot 43$ with $N_{x^6}=51840$ and $R_6=0.0018$
$N=35797223=7\cdot 11\cdot 17\cdot 23\cdot 29\cdot 41$ with $N_{x^6}=408240$ and $R_6=0.011$
$N=28527037=7\cdot 11\cdot 17\cdot 19\cdot 31\cdot 37$ with $N_{x^6}=18109$ and $R_6=0.000635$

Answer 1

1. Yes, for square free $N$ the formula is $$\prod_{i=1}^d\left(1+\frac{p_i-1}{(d,p_i-1)}\right)$$

2. The above expression will equal $N$ iff $(d,p_i-1)=1$ for all $i$. For odd $d$ we can find arbitrarily many primes with this property. It follows that the supremum of $R_d$ is 1, for odd $d$ which includes large values of $d$

Conversely, for any given $d$ we can construct $N$ from primes all of which are 1 mod $d$. In such a way we can find $N$ such that $R_d(N)=O(d^{-d})$, but such $N$ are sparse.