The transformation is $$T:M_{\leq n}(\mathbb{C})\rightarrow M_{\leq n}(\mathbb{C})$$ and defined by $$T(A) = A^t-A.$$
If we try to take a basis $B$ and calculate the determinant of $xI-[T]_B$, we get a hard determinant to calculate and I didn't find a way to calculate it. Instead, I noticed that $$m_T(x) = x(x+2)=x^2+2x$$ satisfies $m_T(T)=0$, and therefore it is the minimal polynomial of $T$ (it's easy to check there is no polynomial of the form $x+b$ such that $T+b = 0$).
Now I used the well-known $$f_T(x)|(m_T(x))^{dim(M_{\leq n}(\mathbb{C}))}=x^{n^2}(x+2)^{n^2},$$ where $f_T$ is the characteristic polynomial, and since $deg f_T(x) = n^2$, we get the fact that $f_T(x)$ is of the form $x^k(x+2)^{n^2-k}$. Now we just need to find $k$.
By Cayley Hamilton we get: $$(*) T^k(T+2Id)^{n^2-k}=0.$$
Then I proved using induction that: $$(T+2Id)^m(A) = 2^{m-1}A+2^{m-1}A^t$$ $$T^m(A) = (-1)^m2^{m-1}A+(-1)^{m+1}2^{m-1}A^t$$
Then I plugged it into $(*)$ and got $0$ doesn't matter what $k$ is. So how do you find the characteristic polynomial?
It is enough to compute the dimension of the eigenspace $A^t-A=0$ implies that $A^t=A$ and the dimension of symmetric matrices is ${{n(n+1)}\over 2}$, this implies that the characteristic polynomial is $X^{{n(n+1)}\over 2}(X+2)^{{n(n-1)}\over 2}$