The Cohen reals form a symmetric set but their enumeration is not symmetric

F1 shows that $\{n\}$ supports ${\dot{a}}_n$ and $\mathrm{\varnothing }$ supports $\dot{A}$; their subnames are checks, so both are HS. For $n\mathrm{\ne }m$, below any condition choose a fresh bit coordinate $k$ and set opposite bits at $(n,k),(m,k)$. Hence the set forcing ${\dot{a}}_n\mathrm{\ne }{\dot{a}}_m$ is dense. Every finite collection is therefore forced to have its displayed size, so $A$ is infinite.

Let $\dot{e}$ be the canonical graph $n\mapsto {\dot{a}}_n$. Given finite $E$, choose $n\mathrm{\notin }E$ and $m\mathrm{\notin }E\cup \{n\}$. Their transposition fixes $E$ but sends the graph value at $n$ from ${\dot{a}}_n$ to ${\dot{a}}_m$, so it does not fix $\dot{e}$. Thus no finite $E$ supports that name.

Now let $p⊩\dot{f}:\check{\omega }\to \dot{A}$ be onto and let the finite set $E$ support $\dot{f}$ and contain the first-coordinate support of $p$. Choose $n\mathrm{\notin }E$. Since $p$ forces surjectivity, some $q\le p$ and $k\in \omega$ satisfy $q⊩\dot{f}(\check{k})={\dot{a}}_n$. Choose $m$ outside $E\cup \{n\}$ and outside the first-coordinate support of $q$, and let $\pi$ swap $n,m$. Then $\pi p=p$, $\pi \dot{f}=\dot{f}$, and F2 gives

Because the $m$-coordinate is absent from $q$, $q$ and $\pi q$ agree on their common domain and have a common extension. That extension forces ${\dot{a}}_n={\dot{a}}_m$, contrary to step 1.1. Therefore no HS name can enumerate $A$.