As $A$ is real, and normal, it commutes with $A^\top$, and these matrices can be simultaneously diagonalised by a unitary $U$. That is, both sides of $\det((A+A^\top)/2)=\det A$ can be conjugated by $U$ to obtain $\det(D/2+D'/2)=\det D$, with $D,D'$ diagonal matrices, and multisets of their entries are equal, each entry a positive real.

In the minimal nontrivial case of 2x2 matrices, assuming $A\neq A^\top$, the latter equality becomes $D_1 D_2=(D_1/2+D_2/2)^2$, with $D_2<D_1$, and $D=diag(D_1,D_2)$, $D'=diag(D_2, D_1)$. But this is only possible if $D_1=D_2$, a contradiction showing that $A=A^\top$ in this case. In general, it's not immediately clear how to reach the same conclusion.

As $A$ is real, and normal, it commutes with $A^\top$, and these matrices can be simultaneously diagonalised by a unitary $U$. That is, both sides of $\det((A+A^\top)/2)=\det A$ can be conjugated by $U$ to obtain $\det(D/2+D'/2)=\det D$, with $D,D'$ diagonal matrices, and multisets of their entries are equal, each entry a positive real.

In the minimal nontrivial case of 2×2 matrices, assuming $A\neq A^\top$, the latter equality becomes $D_1 D_2=(D_1/2+D_2/2)^2$, with $D_2<D_1$, and $D=\operatorname{diag}(D_1,D_2)$, $D'=\operatorname{diag}(D_2, D_1)$. But this is only possible if $D_1=D_2$, a contradiction showing that $A=A^\top$ in this case. In general, it's not immediately clear how to reach the same conclusion.

As $A$ is real, and normal, it commutes with $A^\top$, and these matrices can be simultaneously diagonalised by a unitary $U$. That is, both sides of $\det((A+A^\top)/2)=\det A$ can be conjugated by $U$ to obtain $\det(D/2+D'/2)=\det D$, with $D,D'$ diagonal matrices, and multisets of their entries are equal, each entry a positive real.

In the minimal nontrivial case of 2x2 matrices, assuming $A\neq A^\top$, the latter equality becomes $D_1 D_2=(D_1/2+D_2/2)^2$, with $D_2<D_1$, and $D=diag(D_1,D_2)$, $D'=diag(D_2, D_2)$. But this is only possible if $D_1=D_2$, a contradiction showing that $A=A^\top$ in this case. In general, it's not immediately clear how to reach the same conclusion.

As $A$ is real, and normal, it commutes with $A^\top$, and these matrices can be simultaneously diagonalised by a unitary $U$. That is, both sides of $\det((A+A^\top)/2)=\det A$ can be conjugated by $U$ to obtain $\det(D/2+D'/2)=\det D$, with $D,D'$ diagonal matrices, and multisets of their entries are equal, each entry a positive real.

In the minimal nontrivial case of 2x2 matrices, assuming $A\neq A^\top$, the latter equality becomes $D_1 D_2=(D_1/2+D_2/2)^2$, with $D_2<D_1$, and $D=diag(D_1,D_2)$, $D'=diag(D_2, D_1)$. But this is only possible if $D_1=D_2$, a contradiction showing that $A=A^\top$ in this case. In general, it's not immediately clear how to reach the same conclusion.