My attempt:
Suppose that $\{e_1,\dots,e_n\}$ is an orthonormal basis and $L(e_i)=a_{ik}e_k$ (Einstein sommation). Then, $(a_{ik})$ is a symmetric matrix (by given information).
Now consider another orthonormal basis $\{f_1,\dots,f_n\}$. We can find a matrix $(b_{ij})$ such that $f_i=b_{ij}e_j$. Then, $$ L(f_i)=c_{ik}f_k=c_{ik}b_{kj}e_j.$$ Since $L$ has a symmetric matrix representation w.r.t. $\{e_1,\dots,e_n\}$, we now that $(c_{ik}b_{kj})$ is symmetric (and a product of two matrices). How I can show that $(c_{ik})$ has to be symmetric?
The matrix $(m_{ij})$ given by $m_{ij} = c_{ik}b_{kj}$ is the matrix of a transformation with respect to two different bases. In particular, we have $L(f_i) = m_{ij}e_j$. As such, $(m_{ij})$ is not necessarily symmetric.
We can make this a matrix of $L$ with respect to one basis as follows. Note that $$ L(f_i) = L(b_{i \ell} e_\ell) = b_{i \ell} L(e_{\ell}) = m_{ij}e_j \implies\\ L(e_\ell) = b^{-1}_{i \ell} m_{ij} e_j = b^{-1}_{i \ell} c_{ik} b_{kj} e_j. $$ In other words, we have $a_{ij} = b_{p i}^{-1}c_{p q} b_{q j}$. Or, solving for $c_{ij}$, we have $$ c_{ij} = b_{i p} a_{p q}b^{-1}_{j q}. $$ Now, note that because $e_i,f_i$ are both orthonormal bases, we have $b^{-1}_{ij} = b_{ji}$, so that the above becomes $$ c_{ij} = b_{i p} a_{p q}b_{q j}. $$ Conclude that $(c_{ij})$ is symmetric.