I think one way of explaining this is via quadratic forms. The usual correspondence sends $X$ to the quadratic form $X(v)=vXv^t,$ $v$ a row vector. But in characteristic $2$, this formula simplifies to
$$X(v)=(vX_0)^2.$$
Now examine what happens to $AXA^t$. For any characteristic, we get $AXA^t(v)=vAXA^tv^t=(vA)X(vA)^t=X(vA).$ In characteristic $2$, this implies that $$(v(AXA^t)_0)^2=(vA(X_0))^2,$$
so
$$(AXA^t)_0=A(X_0)$$
as desired.