Write $A = B + C$ where $B$ is symmetric and $C$ is antisymmetric. The assumption is that $B$ is positive-definite and $\det(A) = \det(B)$. we wish to prove $C=0$.
First, every symmetric positive-definite matrix is of the form $g \cdot g^t$ for some $g \in \mathrm{GL}_n(\mathbf{R})$. Thus, by acting by this group, we can assume $B =1$. In more detail, if $B = g \cdot g^t$, take the equation $A = B + C$ and multiply on the left by $g^{-1}$ and on the right by $g^{-t}$. Let $A_1 = g^{-1} A g^{-t}$ and $C_1 = g^{-1} C g^{-t}$. Then $A_1 = 1 + C_1$. Observe that $C_1$ is still anti-symmetric. Moreover, because $\det(A) = \det(B)$, $\det(A_1) = 1$.
Now, as $C_1$ is anti-symmetric, by the spectral theorem I can assume $C_1$ is block diagonal with two-by-two anti-symmetric blocks. (See the Wikipedia page for "Skew-symmetric matrix"). More specifically, $C_1 = h C_2 h^{t}$ for some $h \in O(n)$ and $C_2$ block-diagonal with anti-symmetric $2 \times 2$ blocks. Write $A_2 = h^{-1} A_1 h^{-t}$. We have $A_2 = 1 + C_2$, and $\det(A_2) = 1$.
The determinant of $1 + C_2$ is now immediate to compute, and one gets $C_2 = 0$ from the equality $1 = \det(1+C_2)$. As $C_2 = 0$, $C_1 = 0$ and thus $C=0$.