Denote scalar product of vectors $v,u$ by $(v,u)$, norm of vector $v$ by $\|v\|=\sqrt{(v,v)}$.
Lemma 1. Let $A$ be a symmetric positive operator on $\mathbb{R}^n$, $f\in \mathbb{R}^n$ be a vector. Then $(Af,f)\cdot (A^{-1} f,f)\geq \|f\|^4$.
Proof. Let $A=B^2$, where $B=\sqrt{A}$ is positive. Then $(Af,f)=(B^2f,f)=(Bf,Bf)=\| Bf\|^2$, $(A^{-1}f,f)=\|B^{-1}f\|^2$ and we have to prove $\|Bf\|\cdot \|B^{-1} f\|\geq \|f\|^2$, but, by the Cauchy-Schwarz inequality, $\|Bf\|\cdot \|B^{-1} f\| \geq (Bf, B^{-1} f)=(f,f)=\|f\|^2$, as desired.
Lemma 2. If for some positive symmetric matrix $A$ diagonal elements are equal to 1, then for $A^{-1}$ diagonal elements are not less then 1.
Proof. Apply Lemma 1 to basis vectors.
Now assume that $U:=V^TSV={\rm diag}(c_1,\dots,c_n)$. Then ${\rm tr}\, V^T SV=\sum c_i$,
$$
{\rm tr}\, S={\rm tr}\, V^TS(V^T)^{-1}={\rm tr}\, V^TSV (V^T V)^{-1}={\rm tr}\, U F^{-1},
$$
where $F=V^TV$ is a symmetric positive matrix with unit diagonal elements. So, by Lemma 2 diagonal elements $w_1,\dots,w_n$ of $F^{-1}$ are not less than $1$ and so ${\rm tr}\, S=\sum c_i w_i\geq \sum c_i={\rm tr}\, U$.
The matrices $\pmatrix{1&3\cr0&2\cr}$ and $\pmatrix{1&0\cr0&2\cr}$ are similar, so there is a change of basis that transforms one into the other, but one is symmetric and the other is not, so, yes, there are transformations that have a symmetric matrix with respect to one basis and not to another basis.
Best Answer
I interpret the question as follows:
Let $f$ be a symmetric bilinear form on a finite dimensional vector space over a field $K$ whose characteristic in not two and which has at least four elements, and let $A$ be the matrix of $f$ with respect to some basis $B$ of $V$. Prove or disprove: The eigenvalues of $A$ are independent of the choice of $B$.
We disprove the statement as follows.
Putting $$ V:=K,\quad f(x,y):=xy,\quad B:=\{b\}, $$ we get $A=(b^2)$. In view of the assumptions on $K$, we can choose nonzero $b$ and $b'$ in $K$ so that $b^2\neq b'^2$.