A linear transformation such that its matrix is equal to its transpose with respect to each basis Suppose that I have a linear transformation $f:V\to V$ such that for every basis $E=\lbrace e_1,\ldots,e_n \rbrace$ of $V$ the matrix representation $[f]_E^E$ of $f$ with respect to $E$ is equal to its transpose (i.e., $[f]_E^E = ([f]_E^E)^T$). Does this imply that $f$ is a multiple of the identity function? (i.e., $f = \alpha I$ for some $\alpha \in \mathbb{R}$). 
 A: Yes. This would mean that $f$ is self-adjoint with respect to any inner product on $V$ and this in turn would imply that $f$ is a multiple of the identity. For more details, see this answer. 
A: Here is a relatively ignorant solution.  If $A$ is a matrix for $f$ with respect to a basis, then the set of matrices for $f$ with respect to all bases is the same as the set of matrices similar to $A$.  Hence if all matrices for $f$ are symmetric, then $A$ is a symmetric matrix such that $SAS^{-1}$ is also symmetric for all invertible matrices $S$. So for each invertible $S$,
$$SAS^{-1}=(SAS^{-1})^T=(S^{-1})^T A^T S^T=(S^{-1})^TAS^T.$$
Multiplying on the left by $S^T$ and on the right by $S$, this implies $$S^T SA = AS^TS,$$
that is, $A$ commutes with all invertible matrices of the form $S^TS$. Let $S$ be a diagonal matrix with all positive distinct diagonal entries.  Then $S^TS$ commuting with $A$ implies that $A$ is a diagonal matrix. This conclusion follows from what is obtained immediately by writing down what the product on each side looks like when $S^TS$ is diagonal with distinct diagonal entries.
Consider $S=I+\begin{bmatrix}1&1&0&\cdots\\
1&1&0&\cdots\\
0&0&0&\cdots\\
\vdots&\vdots&\vdots&\ddots\end{bmatrix}$ (all $0$s in the ellipses).  Then the diagonal matrix $A$ commuting with $S^TS=S^2=I+\begin{bmatrix}4&4&0&\cdots\\
4&4&0&\cdots\\
0&0&0&\cdots\\
\vdots&\vdots&\vdots&\ddots\end{bmatrix}$ implies that the first two diagonal entries of $A$ are equal, again using nothing other than multiplying both ways and seeing what you get.  By repeating with that $2$-by-$2$ block of $1$s sliding down the diagonal, we conclude that all of the diagonal entries of $A$ are equal, so $A$ is a scalar multiple of the identity matrix.  
