# Just want to confirm: Do two CNOT gates cancel each other?

I see somewhere that this happens: But I wonder if this is just identity.

Yes, it is. If the bottom qubit is 0, neither gate does anything to the top qubit. If the bottom qubit is 1, both gates apply $X$. But since $X^2=\mathbb{I}$, the net effect is that nothing happens. Hence, overall, nothing happens.
Another way to see this is to look at the unitary matrix of controlled-not. $$\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right)$$ It’s reasonably easy to see that the eigenvalues are 1,1,1,-1 (evidently, $|00\rangle$ and $|01\rangle$ are +1 eigenvectors, leaving behind a $2\times 2$ matrix like Pauli $X$, which we know has $\pm 1$ eigenvalues), so the square obviously has eigenvalues 1,1,1,1 and the only 4x4 unitary matrix with all ones eigenvalues is the identity matrix.
Equally, direct calculation: $$\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right)\cdot \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right)=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$$
Any Hermitian gate is "self-canceling". Proof: since any gate $U$ is unitary $$UU^{\dagger}=U^{\dagger}U=I$$ If $U$ is also Hermitian, $U=U^{\dagger}$ and $$UU=I$$
CNOT gate $$\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right)$$ is Hermitian by inspection.