Those are examples of palindromic or anti-palindromic polynomials (I will call them pal and a-pal respectively [and (a)pal, for one that is either] in this answer). The identity matrix of order $n$ has characteristic polynomial $(x - 1)^n$, whose expansion has binomial coefficients, which are symmetric. This makes it obvious why its characteristic polynomial is (a)pal.
Now more generally, a polynomial is (a)pal iff all its roots are multiplicatively symmetric about $1$. That is, for each root $\lambda$ of a polynomial that is (a)pal, $\dfrac 1 \lambda$ is also a root.
One obvious and simple implication of this is that zero can never be a root, and any matrix $A$ with (a)pal characteristic polynomial is therefore non-singular. Since the eigenvalues of the inverse matrix $A^{-1}$ are exactly the reciprocals of the eigenvalues of $A$, this also implies that $A$ and $A^{-1}$ have the same spectrum.
We can do better than this.
Theorem
A square matrix with entries from an algebraically closed field has (a)pal characteristic polynomial if and only if it is similar to its inverse.
Proof: Let $A$ be a matrix with (a)pal characteristic polynomial, and let $$J = P^{-1}AP = \operatorname{diag}(J_1, \ldots, J_p)$$
be its Jordan normal form. Each block $J_k$ is of the form
\begin{equation*}
J_k = \begin{bmatrix}
\lambda_k & 1 & 0 & \cdots & 0\\
0 & \lambda_k & 1 & \cdots & 0\\
0 & 0 & \lambda_k & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & \lambda_k
\end{bmatrix}.
\end{equation*}
We know that $A^{-1}$ has Jordan normal form $J^{-1} = P^{-1}A^{-1}P$, where $J^{-1} = \operatorname{diag}(J_1^{-1}, \ldots, J_p^{-1})$, and each block $J_k^{-1}$ is of the form
\begin{equation*}
J_k^{-1} = \begin{bmatrix}
\frac 1 {\lambda_k} & 1 & 0 & \cdots & 0\\
0 & \frac 1 {\lambda_k} & 1 & \cdots & 0\\
0 & 0 & \frac 1 {\lambda_k} & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & \frac 1 {\lambda_k}
\end{bmatrix},
\end{equation*}
and has the same size as $J_k$. However, as the characteristic polynomial of $A$ is (a)pal, $J_k^{-1} = J_l$, for some $l$ (and hence $J_l^{-1} = J_k$). Thus, $J$ and $J^{-1}$ differ only by a rearrangement of the blocks $J_1, \ldots, J_p$, and are therefore similar. Specifically, there exists a permutation matrix $Q$ such that $J^{-1} = QJQ^{-1}$ (see note below). Then
\begin{equation*}
A^{-1} = PJ^{-1}P^{-1} = P(QJQ^{-1})P^{-1} = (PQP^{-1})(PJP^{-1})(PQP^{-1})^{-1} = RAR^{-1},
\end{equation*}
where $R = PQP^{-1}$. Thus, if $A$ has (a)pal characteristic polynomial, it is similar to its inverse.
The converse follows immediately by observing that similar matrices have the same characteristic polynomial. $\qquad \square$
Note: The permuting matrix $Q$ for transforming $J$ to $J^{-1}$ can be obtained by writing the identity matrix $I$ of the same order as $J$ in the block diagonal form $I = \operatorname{diag}(I_1, \ldots, I_p)$, where $I_k$ is the identity matrix of the same order as $J_k$, and then applying the same permutation on these respective blocks as applied on the blocks of $J$ to obtain $J^{-1}$. That is, if $$J^{-1} = \operatorname{diag}(J_{\sigma(1)}, \ldots, J_{\sigma(p)}),$$ where $\sigma$ denotes the permutation applied to the blocks, then $$Q = \operatorname{diag}(I_{\sigma(1)}, \ldots, I_{\sigma(p)}).$$
