A symmetric and transitive relation on a two-element set that is not reflexive on it

$u\mathrm{\ne }v$, because $v$ has the element $u$ and $u$ has none; so $A$ has exactly the two elements $u$ and $v$.

$R$ is a relation on $A$: its only element is the ordered pair $(u,u)$, and both coordinates lie in $A$, so $R\subseteq A\times A$.

The three characteristic sets of $R$ are $\mathrm{dom}R=\mathrm{ran}R=\mathrm{fld}R=\{u\}$, since $(u,u)$ is its only pair.

$R$ is symmetric: the only pair in $R$ is $(u,u)$, whose reversal is itself. It is transitive: the only composable pair of members is $(u,u)$ with $(u,u)$, and the required conclusion $(u,u)\in R$ holds.

$R$ is reflexive on $\mathrm{fld}R$: the only element of $\{u\}$ is $u$, and $(u,u)\in R$.

$R$ is not reflexive on $A$: $v\in A$, and $(v,v)\mathrm{\ne }(u,u)$ because $v\mathrm{\ne }u$, so $(v,v)\mathrm{\notin }R$. The failure is therefore confined to the single element of $A$ lying outside $\mathrm{fld}R$; symmetry and transitivity say nothing about such a point, which is exactly why they do not imply reflexivity on $A$.