Proving unsatisfiability of a propositional formula
If we take the problem precisely as stated, the answer is negative, assuming $P\neq NP$. Indeed, checkSymmetry is trivially implementable in polynomial time, since it takes as input both the formula $F$ and the assignment $A$, so it can check those. Hence, if a reduction existed, that would imply that unsatisfiability is polynomially decidable, hence $P=NP$.



(Update: my guess below was not what the OP wanted, but I am leaving it here anyway.)

However, I guess your checkSymmetry is meant to take only $F$ as input, and check something equivalent to the validity of $F \iff \bar F$, where $\bar F$ is $F$ with every atomic formula negated.

Then, we can reduce checking validity of any propositional formula to checking symmetry.

Given any $F$, take $G := P \land \lnot F$, where $P$ is an atomic formula not occuring in $F$. We have
$$
\models F \iff \models (G \iff \bar G)
$$

($\Rightarrow$)
Assume $\models F$: we deduce immediately $\models \bar F$.

We need to prove that
$$
\models (P \land \lnot F) \iff (\lnot P \land \lnot \bar F)
$$
but we know $F$ and $\bar F$ are true for all assignments, so the above simplifies to $\models \bot \iff \bot$ which is trivial.

($\Leftarrow$) Assume $\models G \iff \bar G$. 
We need to prove $\models F$.
Let $A$ be an assignment for $F$. Extend it to $B:=A[P\mapsto \top]$.
We have
$$
B \models G \iff \bar G
$$
which is
$$
B \models (P \land \lnot F) \iff (\lnot P \land \lnot \bar F)
$$
which, thanks to the choice of $B(P)$ simplifies to
$$
B \models \lnot F \iff \bot
$$
hence $B \models F$, and we also have $A \models F$ since $P$ does not occur in $F$.