Question: Let $R$ be a PID and let $n \geq 2$ be even. What is $H^0_{\operatorname{fppf}}(R,\operatorname{SL}_n/\mu_2)$?
Attempt at an Answer: From the short exact sequence of algebraic groups $1 \to \mu_2 \to\operatorname{SL}_n\to\operatorname{SL}_n/\mu_2 \to 1$, we obtain a long exact sequence in fppf cohomology given by $$1 \longrightarrow H^0_{\operatorname{fppf}}(R, \mu_2) = \{\pm 1\} \longrightarrow H^0_{\operatorname{fppf}}(R, \operatorname{SL}_n) = \operatorname{SL}_n(R) \longrightarrow H^0_{\operatorname{fppf}}(R,\operatorname{SL}_n/\mu_2) \longrightarrow H^1_{\operatorname{fppf}}(R,\mu_2) = R^\times/R^{\times 2} \longrightarrow H^1_{\operatorname{fppf}}(R,\operatorname{SL}_n) = 1.$$
Let $K$ be the fraction field of $R$. From the above, I would guess that $H^0_{\operatorname{fppf}}(R,\operatorname{SL}_n/\mu_2)$ may be described as the subgroup of $\operatorname{SL}_n(\overline{K})/\{\pm 1\}$ consisting of matrices $M$ whose entries are units in a quadratic extension of $K$ such that the Galois-conjugate of $M$ is given by $-M$. However, I do not know how to prove this. I guess I could try to show that this description is the sheafification of $\operatorname{SL}_n(R)/\{\pm 1\}$, but that seems to require guessing the answer first, which is unsatisfying.
I do know how to answer this question when $R$ is replaced by $K$ and fppf cohomology is replaced by Galois cohomology, but I'm not sure how to imitate that argument over $R$.
Edit: I had previously asked this question with $\mathbb{G}_m$ replacing $\operatorname{SL}_n$; as Mindlack pointed out in the comments, the group $H^0_{\operatorname{fppf}}(R,\mathbb{G}_m / \mu_2)$ can be easily computed using the Kummer sequence.