In Short: Assuming one-way permutations exist, can we construct one that has no trapdoor?
More info:
A one-way permutation is a permutation $\pi$ which is easy to compute, but hard to invert (see the one-way-function tag wiki for a more formal definition). We usually consider families of one-way permutation, $\pi = \{\pi_n\}_{n \in \mathbb{N}}$, where each $\pi_n$ is a one-way permutation, acting on a finite domain $D_n$. A trapdoor one-way permutation is defined as above, except that there exists a trapdoor set $\{t_n\}_{n \in \mathbb{N}}$ and an poly-time inverting algorithm $I$, such that for all $n$, $|t_n| \le {\rm poly}(n)$, and $I$ can invert $\pi_n$ provided that it is given $t_n$.
I know one-way permutations which are generated so that it is infeasible to find the trapdoor (yet the trapdoor exists). An example, based on RSA-assumption, is given here. The question is,
Do there exist (families of) one-way permutations which do not have a trapdoor (set)?
Edit: (More Formalization)
Assume there exists some one-way permutation $\pi$ with (infinite) domain $D \subseteq \{0,1\}^*$. That is, there exists a probabilistic polynomial-time algorithm $\mathcal{D}$ (which, on input $1^n$, induces some distribution over $D_n=\\{0,1\\}^n \cap D$), such that for any polynomial-time adversary $\mathcal{A}$, any $c>0$, and all sufficiently large integer $n$:
$\Pr[x \leftarrow \mathcal{D}(1^n) \colon \quad \mathcal{A}(\pi(x))=x]<n^{-c}$
(The probability is taken over the internal coin tosses of $\mathcal{D}$ and $\mathcal{A}$.)
The question, is whether we can construct a one-way permutation $\pi'$, for which there exists a probabilistic polynomial-time algorithm $\mathcal{D}'$ such that for any poly-size family of circuits $\mathcal{A}'=\{\mathcal{A}'_n\}_{n \in \mathbb{N}}$, any $c>0$, and all sufficiently large integer $n$:
$\Pr[x \leftarrow \mathcal{D}'(1^n) \colon \quad \mathcal{A}'_n(\pi'(x))=x]<n^{-c}$
(The probability is taken over the internal coin tosses of $\mathcal{D}'$, since $\mathcal{A}'$ is deterministic.)