Though not an exact duplicate (as far as I can tell), this question is very similar to the following existing questions on this site, about the minimal polynomials of $A\mapsto A^t-A$, of $A\mapsto A-2A^t$, and of $A\mapsto-A^t$. In each case the question is about a linear combination of the transposition map (from square matrices to themselves) and the identity map $I$ (on square matrices). Since for eigenvalue problems, linear combinations with the identity map are easily handled (and more generally linear combinations of commuting operators) it is a good idea to start with the transposition map $\tau:A\mapsto A^t$.
It is clear that $\tau$ is an involution (that is, $\tau^2=I$), and it follows that the minimal polynomial divides $X^2-1=(X+1)(X-1)$. As the latter two factors are relatively prime (the real numbers do not have characteristic$~2$), $\tau$ is diagonalisable, and its eigenvalues are contained in the set $\{-1,1\}$ Indeed the eigenspace for $\lambda=-1$ is the set of antisymmetric matrices, which has positive dimension when $n>1$, and the eigenspace for $\lambda=1$ is the set of symmetric matrices, which has positive dimension when $n>0$. So the minimal polynomial of $\tau$ is $1$ when $n=0$, it is $X-1$ when $n=1$, and it is $(X+1)(X-1)=X^2-1$ as soon as $n\geq2$. For the characteristic polynomial you must raise each factor $X-\lambda$ to a power equal to the dimension of the eigenspace for$~\lambda$, which is easily computed.
Now for a linear combination $T=aI+b\tau$ with $b\neq0$, the eigenspaces do not change; the only thing that is needed is to replace the eigenvalue $\lambda$ by $a+b\lambda$ (because the term $aI$ acts by a scalar $a$, and the term $b\tau$ by a factor $b\lambda$, on the eigenspace of$~\tau$ for $\lambda$). In the polynomials one must similarly replace the factors $X-\lambda$ by $X-(a+b\lambda)$ as only change (the condition $b\neq0$ ensures the two factors remain relatively prime). In your case ($a=1=b$) you get eigenvalues $0$ from $\lambda=-1$ (antisymmetric matrices) and $2$ from $\lambda=1$ (symmetric matrices). The other linked cases are entirely similar.