Is it possible to find two sets such that their hash xor summation is the same? Consider two sets $A=\{a_1,a_2,\cdots,a_n\}, B=\{b_1,b_2,\cdots,b_m\}$; $m,n$ can be different. we can calculate the xor summation of each element's hash value: $XOR(A)=(hash(a_1)\oplus hash(a_2) \oplus \cdots \oplus hash(a_n))$ and $XOR(B)=(hash(b_1) \oplus hash(b_2) \oplus \cdots \oplus hash(b_m))$. Is it possible that set $A$ does not equal set $B$, yet $XOR(A)=XOR(B)$?
As for the feasibility of finding such a colliding set, well, as long as the sets are allowed to be relatively long (e.g. at least as long as it bit values being xor'ed), then it is easy.

First off, we assume that the hash output is $n$ bits.  Then, we select $n+1$ distinct values $X_i$ and set $\delta_i = Hash(X_i)$.  Then, we use linear algebra to find a subset of the values $\{ \delta_1, \delta_2, ..., \delta_n \}$ that xor to $\delta_0$ (note, there is a probability that there won't be such a subset; that probability can be minimized by including a few extra values).

With such a subset $\{ \delta_{z_0}, \delta_{z_1}, ..., \delta_{z_k} \}$, we have $$Hash(X_0) = Hash(X_{z_0}) \oplus Hash(X_{z_1}) \oplus ... \oplus Hash(X_{z_k})$$

That is, the sets $\{ X_0 \}$ and $\{X_{z_0}, X_{z_1}, ..., X_{z_k} \}$ collide