Show that $det(\phi_A)=(detA)^{n+1}$ 
$E=M_n(\mathbb{R})$, $F=S_n(\mathbb{R})$. (EDIT: symmetric matrices)
Let $A\in E$ fixed. Denote $\phi_A:F\rightarrow F$, 
  $S→ASA^{T}$.
Show that $det(\phi_A)=(detA)^{n+1}$.

My attempt :
Let $S$ an eigenvector of $\phi_{A}$ associated to  $\mu$ and $X$ an eigenvector of $A^{T}$  associated to $\lambda$ : 
$$AS A^{T}X = \mu SX$$ then $\lambda ASX=\mu SX.$$
Therefore if $\lambda \ne 0$ and $SX\ne0$ we have $\frac{\mu}{\lambda}$ is a eigenvalue of $A$
If $A$ is diagonalizable, let $(U_ {i})$ be a basis of eigenvectors of $A$ (associated with $(\lambda_{i})$) then $A=PDP^{-1}$.
Then  $(A^{T})^{T}=(P^{-1})^{T}DP$. So columns vectors $(P^{-1})^{T}$ form a basis of eigenvectors of $A^{T}$. Note that $(V_ {i})$.
We know that $A$ and its transpose have the same eigenvalues ​​with the same multiplicity.
If we denote $(E_ {i})$ the canonical basis of $\mathbb{R}^n$, we have $E_{i}=PU_{i}$ and $E_{i}= (P^{-1})^{T}V_{i}$. This gives us $U_{i}=P ^{-1} (P^{-1})^{T}V_ {i}$
Now I am stuck.
Thank you in advance for your time.
 A: Consider the inner product in $S_n(\mathbb{R})$ as $\langle X,Y\rangle=tr(XY)$.
Prove that $\phi_{A}^t$ with respect to this inner product is $\phi_{A^t}$.
Hint: $\langle \phi_A(X),Y\rangle=\langle X,\phi_A^t(Y)\rangle$.
Now $\phi_A\circ\phi_{A^t}=\phi_{AA^t }$. Remind that $AA^t$ has an orthonormal basis of eigenvectors $v_1,\ldots,v_n$ associated to the eigenvalues $\lambda_1,\ldots,\lambda_n$
respectively.
Prove that the eigenvalues of $\phi_{AA^t}$ are $\lambda_{i}\lambda_j$. 
Hint: Consider the matrices $v_iv_i^t$ and $v_iv_j^t+v_jv_i^t$ for $ i,j\in \{1,\ldots,n\}$ and $i\neq j$. Notice that these matrices form a basis of  $S_n(\mathbb{R})$.
Prove $\det(\phi_{AA^t})=det(AA^t)^{n+1}$. Hint: Determinant is the product of eigenvalues. 
Thus, since $\phi_{A}^t=\phi_{A^t}$ and $det(\phi_A)=\det(\phi_A^t)$then $$\det(\phi_{AA^t })=\det(\phi_A\circ\phi_{A^t})=\det(\phi_A)\det(\phi_{A^t})=\det(\phi_A)\det(\phi_{A}^t)=\det(\phi_A)^2$$
Finally, $\det(\phi_A)^2=det(AA^t)^{n+1}=\det(A)^{2(n+1)}$ and $det(\phi_A)=\det(A)^{n+1}$.
