For a random permutation $P$, what's the probability of the following event? For a random permutation $P$ and $q$ distinct inputs $x_1,\ldots,x_q\in\{0,1\}^n$, what's the probability of the event that there exists at least one collision among $\{P(x_1)\oplus x_1,\ldots,P(x_q)\oplus x_q\}$?

Note: I have tried to compute this probability, but it is quite hard for me to obtain the result, either for the exact probability or the lower bound of this probability. The following is my attempt to solve this problem when $q=2,3$.

Denote by $\mathsf{E}$ the event that there exists at least one collision among $\{P(x_1)\oplus x_1,\ldots,P(x_q)\oplus x_q\}$.


If $q=2$, it is easily seen that $\Pr[\mathsf{E}]=\Pr[P(x_1)\oplus x_1=P(x_2)\oplus x_2]=\frac{1}{2^n-1}$

If $q=3$, we have 3 random elements $P(x_1)\oplus x_1,P(x_2)\oplus x_2,P(x_3)\oplus x_3$. We focus on computing $\overline{\mathsf{E}}$ the complementary event of $\mathsf{E}$:
 \begin{align}\Pr[\overline{\mathsf{E}}]=&\Pr[P(x_1)\oplus x_1\neq P(x_2)\oplus x_2 \wedge P(x_1)\oplus x_1 \neq P(x_3)\oplus x_3 \wedge P(x_2)\oplus x_2 \neq P(x_3)\oplus x_3]\\
=&\Pr[P(x_1)\oplus x_1 \neq P(x_3)\oplus x_3 \wedge P(x_2)\oplus x_2 \neq P(x_3)\oplus x_3\mid P(x_1)\oplus x_1\neq P(x_2)\oplus x_2]\cdot\Pr[P(x_1)\oplus x_1\neq P(x_2)\oplus x_2]

\end{align}
Without loss of generality, assume that the value of $P(x_1)$ is fixed. There are $2^n-1$ possibilites for $P(x_2)$. Conditioned on the event $P(x_1)\oplus x_1\neq P(x_2)\oplus x_2$ happening, the probability of the event $P(x_1)\oplus x_1\neq P(x_3)\oplus x_3\wedge P(x_2)\oplus x_2\neq P(x_3)\oplus x_3$ may change according to the choices of $P(x_2)$:

If $P(x_2)=P(x_1)\oplus x_1\oplus x_3$ or $P(x_2)=P(x_1)\oplus x_2\oplus x_3$, then the probability of the event $P(x_1)\oplus x_1\neq P(x_3)\oplus x_3\wedge P(x_2)\oplus x_2\neq P(x_3)\oplus x_3$ is $\frac{2^n-2-1}{2^n-2}$.
If $P(x_2)\neq P(x_1)\oplus x_1\oplus x_3$, $P(x_2)\neq P(x_1)\oplus x_2 \oplus x_3$ and further $P(x_2)\neq P(x_1)\oplus x_1\oplus x_2$, then the probability of the event $P(x_1)\oplus x_1\neq P(x_3)\oplus x_3\wedge P(x_2)\oplus x_2\neq P(x_3)\oplus x_3$ is $\frac{2^n-2-2}{2^n-2}$.

Hence,
\begin{align}
\Pr[\overline{\mathsf{E}}]=&amp;\frac{2}{2^n-1}\cdot\frac{2^n-3}{2^n-2}+\frac{2^n-1-3}{2^n-1}\cdot\frac{2^n-4}{2^n-2}\\
=&amp;\frac{2^{2n}-6\cdot 2^n+10}{(2^n-1)(2^n-2)},
\end{align}
and $\Pr[\mathsf{E}]=1-\Pr[\overline{\mathsf{E}}]=1-\frac{2^{2n}-6\cdot 2^n+10}{(2^n-1)(2^n-2)}$.
For $q\ge4$, the computation is complicated and I haven't found a good method to tackle it.
Assuming the $x_i$ are chosen independently of $P$, and the witdth $k$ of $x$ is not too narrow, it makes sense to use the approximation that $x\mapsto P(x)\oplus x$ behaves like a random function, at least as far as collisions are concerned.

Note: I wish I had a proof or convincing argument of that. The best I have right now is that without knowledge of $P$, we need nearly $2^{k/2}$ queries to $x\mapsto P(x)\oplus x$ in order to get a significant advantage distinguishing that from a random function. But that argument does not explain why the approximation still holds for e.g. $k=10$, $q=100$, when it does.

If further assuming that the $x_i$ are distinct, the problem becomes a straightforward application of the birthday problem for cryptographic hashing.