Finding the characteristic polynomial of the $T: \mathcal{M}_{n} (\mathbb{R}) \to \mathcal{M}_{n} (\mathbb{R})$ given by $T(M)=M^{\text{tr}}$

$T^2=Id$ implies the minimal polynomial is $X^2-1=0$ since $T$ is different of the $M\rightarrow -M$ map and the identity, its minimal polynomial cannot be $X$ or $X-1$.

The eigenspace associated to $1$ is the space of symmetric matrices and the eigenspace associated to $-1$ is the space of antisymmetric matrices.

The characteristic polynomial is $(X-1){{n(n+1)}\over 2}(X+1)^{{n(n-1)}\over 2}$ since the space of symmetric matrices has dimension ${{n(n+1)}\over 2}$ and the space of antisymmetric matrices has dimension ${{n(n-1)}\over 2}$ and you can compute the characteristic polynomial of $T$ by decomposing $M(n,\mathbb{R})=Sym(n,\mathbb{R})\oplus Antisym(n,\mathbb{R})$.

$\newcommand{\tr}[0]{\text{tr}}$The answer by Tsemo Aristide is crystal clear. Let me add a couple of comments. Assume $n > 1$.

Given any $M \in \mathcal{M}_{n} (\mathbb{R})$, write $$ M = \frac{1}{2} (M + M^{\tr}) + \frac{1}{2} (M - M^{\tr}). $$ Here the first summand is a symmetric matrix ($N = N^{\tr}$) and the second summand is anti-symmetric ($N = -N^{\tr}$), so that $$ \begin{cases} T(\frac{1}{2} (M + M^{\tr})) = \frac{1}{2} (M + M^{\tr})\\ T(\frac{1}{2} (M - M^{\tr})) = - \frac{1}{2} (M - M^{\tr})\\ \end{cases} $$ We thus have that $\mathcal{M}_{n} (\mathbb{R})$ is the direct sum of the space $\mathfrak{S}$ of symmetric matrices, on which $T$ acts as the identity matrix, and the space $\mathfrak{A}$ of anti-symmetric matrices, on which $T$ act acts as minus the identity. Hence the minimal polynomial of $T$ is $x^{2} - x$, the eigenspaces are $\mathfrak{S}$ for the eigenvalue $1$ and $\mathfrak{A}$ for the eigenvalue $-1$, and $T$ is clearly diagonalizable.

As to the characteristic polynomial, note that $\mathfrak{S}$ has dimension $$ s = \dbinom{n+1}{2}, $$ and $\mathfrak{A}$ has dimension $$ a = \dbinom{n}{2}, $$ so that the characteristic polynomial is $$ (x - 1)^{s} (x + 1)^{a}. $$