Involution action on Brauer group of an abelian variety

Question

Let $k$ be an algebraically closed field of characteristic $p>2$, let $A/k$ be an abelian variety. Let $\iota\colon A\to A, a\mapsto -a$ be the natural involution. Let $x\in\mathrm{Br}(A)[p]$ be a Brauer class on $A$ of order $p$. Is it necessarily true that $\iota^*x=x$?

Answer 1

$\newcommand{\bG}{\mathbb{G}}$Let $X$ be any smooth scheme over an algebraically closed field $k$ of characteristic $p$. From the short exact sequence $0\to\mu_p\to \bG_m\to\bG_m\to 0$ of sheaves on the flat site of $X$ we get $0\to (\mathrm{Pic}\,X)/p\to H^2_{fl}(X,\mu_p)\to H^2_{fl}(X,\bG_m)[p]\to 0$.

Let now $X$ be an abelian variety. Ogus showed in Proposition 1.2 of "Supersingular K3 crystals" that under certain assumptions on a smooth proper variety (that are satisfied for abelian varieties) there is a canonical inclusion $$H^2_{fl}(X,\mu_p)\hookrightarrow H^2_{\mathrm{dR}}(X/k) $$

The map is induced by the identification $H^2_{fl}(X,\mu_p)\simeq H^1_{et}(X,\bG_m/(\bG_m)^p)$ and the logarithmic derivative $\bG_m/(\bG_m)^p\xrightarrow{dlog}Z^1\to \Omega^{\bullet}[1]$. Since there is a canonical isomorphism $H^2_{\mathrm{dR}}(X/k)\simeq\Lambda^2 H^1_{\mathrm{dR}}(X/k)$ the multiplication by an integer $[n]:X\to X$ induces multiplication by $n^2$ on $H^2_{\mathrm{dR}}(X/k)$. In particular, the involution $\iota$ acts by identity and hence it does so on $H^2_{fl}(X,\mu_p)$ and $H^2_{fl}(X,\bG_m)[p]=(\mathrm{Br}\,X)[p]$ where the last identification comes from Gabber's theorem.