Notice that we can find a good $N$ such that after knowing $d(Nx)$ then $x$ can uniquely be determined. For this reason, we shall provide two different approaches;

1st approach. Let $p_1 < p_2 < \dots < p_t$ be all the primes less than $100!$ such that $p_1^R > 100!$. Now, for each $i$, choose large enough primes $q_{ij}$ and $1 \le j \le R$ and put $N = p_1^{\alpha_1} \cdots p_t^{\alpha_t}$ such that $a_i \equiv -j \pmod{q_{ij}}$, $j = 1, \dots, R$. Now, if $d(Nx)$ is divisible by $q_{ij}$ it follows that $v_{p_i}(x) = j$ and this would help us to uniquely determine $x$.

2nd approach. Let us denote by $p_1, \dots, p_t$ the prime divisors dividing $x$. Then, $x = \prod_{i=1}^t p_i^{\alpha_i}$, $\alpha_i \ge 0$. Then if $N = \prod_{i=1}^t p_i^{\beta_i}$ it follows that $d(Nx) = \prod_{i=1}^t (1 + \alpha_i + \beta_i)$. Letting $\beta_i = x^{2^{i-1}}$. Considering $P(x) = \prod_{i=1}^t (1 + \alpha_i + x^{2^i})$ then, $\deg P(x) = 2^t - 1$ and the coefficient of $x^{2^t - 2^{j-1}}$ is equal to $1+\alpha_j$. Take $n > \prod_{i=1}^t (1+\alpha_i)$ then $P(n)$ will be a number in base $n$ that all of its digits can uniquely be determined by $\alpha_i$. So, it remains to choose, $n > \prod_{i=1}^t (1+a_i)$ where $a_i = \max_{i \in A} \alpha_i$ and $\beta_i = n^{2^{i-1}}$.

Now, putting $T = 100!N$ and then $d(\varphi(Tx)) = d(\varphi(100!))Nx$. Plugging $m = p^{S-1}N$ where $t = q \prod_{k=1}^{100!} d(kN)$ for some sufficiently large primes $p$ and $q$. Then
$$\varphi(d(my)) = (q-1)d(Ny)\varphi\left(\prod_{k=1}^{100!} d(kN)\right)$$
It follows that if we consider $\varphi(d(my)) + d(\varphi(Tx)) \pmod{q-1}$ we can determine $d(\varphi(Tx))$ and then determine $x$.