number of queries to recover the permutation

Question

Imagine this game. I pick a permutation $p$ of $1..n$ and give you an oracle. When the oracle is queried with any sequence of $n$ numbers $\in 1..n$, it masks each number by applying some unknown bijection $f$ and then permutes the masked values using $p$. Your goal is to repeatedly query the oracle to recover $p$ with as few queries as possible.

A $\theta(\log(n))$ solution is obvious. First query $1, 2, 3, 4, \ldots$ and $2, 1, 4, 3, \ldots$, the second query reversing each of the $\lfloor \frac{n}{2} \rfloor$ pairs. As swapping a pair in the input swaps the corresponding pair in the output, we identify the $\lfloor \frac{n}{2} \rfloor$ pairs that they map to, albeit with unknown correspondence. (If $n$ is odd, the $n$th item is the one remaining in place so $p(n)$ is known.) Then query $1, 1, 3, 3, \ldots$ to identify which item in each output pair corresponds to the first item of an input pair. With 3 queries we have reduced the problem to one of size $\lfloor \frac{n}{2} \rfloor$ by querying from now on with identical pairs of items; recurse. We use $3\log n \pm O(1)$ queries.

Can we do better? For example, is there a strategy using $\log n \pm O(1)$ queries?

Regarding lower bounds, if $q$ queries is optimal for $\frac{n}{2}$ we can run the strategy twice in parallel, side-by-side, in $q$ queries for $n$ but we still lack information about which $\frac{n}{2}$ of the input maps to the corresponding $\frac{n}{2}$ elements of the output, suggesting we need at least one additional query for $n$. This argument seems to imply $\Omega(\log(n))$ query complexity - can it be made rigorous?