Consider the matrix $$P= \begin{pmatrix} 0 & \ldots & 1 \\ \vdots & 1 & \vdots \\ 1 & \ldots & 0 \end{pmatrix}$$ with $1$s along the "other" main diagonal and $0$s elsewhere. Then $(PA)^t$ is a rotation of the matrix $A$ by $90^\circ$ (you can check this on the basis $E_{i,j}$ of matrices with a 1 in the $(i,j)$ slot and 0s elsewhere). So $A$ is rotatable if and only if $(PA)^t = A$, i.e. $PA=A^t$. Note that $P^2=I$ and $A^tP = A$, so $A^t = AP$ and hence $A$ and $P$ must commute. (For what follows below we assume that $A$ is a real matrix. Otherwise I think all you can say is that $A$ preserves the splitting $\mathbb{C}^n=E_+ \oplus E_-$ as explained below. In the complex case we get symmetric eigenvectors of the $v+Pv$ as well as the skew symmetric vectors $v-Pv$.)
It then follows that $A$ and $A^t$ commute and so, by the Spectral Theorem, $A$ has a unitary basis. Now, notice that $P$ has eigenvalues $1$ and $-1$ with eigenvectors $e_i + Pe_i$ and $e_i-Pe_i$ for $1 \leq i \leq \frac{n+1}{2} $, respectively. Let $E_+ = \ker (P-I)$ and $E_- = \ker (P+I)$. Then $A(E_\pm) \subseteq E_\pm $, so $A$ must preserve the splitting $\mathbb{R}^n = E_+ \oplus E_- $. Note that $\dim E_+ = \left\lfloor \frac{n+1}{2}\right\rfloor$.
Now, the answer to question $2$ is yes if we require the corresponding eigenvalue to be a nonzero real number. If $\lambda$ is an eigenvalue with eigenvector $v \in E_\pm$ and $(\:\: ,\:\: )$ is the standard hermitian inner product on $\mathbb{C}^n$, we have
$$ \lambda(v,v)=(v,A^tv) = (v,APv) = \pm \bar{\lambda}(v,v).$$
So $\lambda \in \mathbb{R}$ if and only if $v \in E_+$ and $\lambda \in i \mathbb{R}$ if and only if $v \in E_-$. Note that $v \in E_+$ if and only if $Pv=v$, which is the same as requiring that $v$ is symmetric. Furthermore, note that the eigenvectors in $E_- \setminus \{0\}$ have complex coefficients as they satisfy $Av=i \mu v$. So it is precisely the real eigenvectors of nonzero eigenvalues that are symmetric. The others will be "skew symmetric".
Question $1$ is a bit trickier, and the answer is no in general. Let us treat the even and odd dimensional cases separately.
Let $n=2m$. If $A$ is rotatable it has the form $$A= \begin{pmatrix} B & B^t P \\ PB^t & PBP \end{pmatrix}$$ for some matrix $B$ (here $P$ and $B$ are matrices of size $m\times m$). Now, $\lambda$ is a real eigenvalue of $A$ with eigenvector $\begin{pmatrix} v \\ Pv \end{pmatrix}$ if and only if $$Bv + B^t v = \lambda v .$$
So $\lambda = 2 \operatorname{Re}(\eta)$, where $\eta$ is an eigenvalue of $B$ and $v$ is an eigenvector of $B+B^t$. Similarly, $\lambda= i \mu$ is a purely imaginary eigenvalue of $A$ with eigenvector $\begin{pmatrix} v \\ -Pv \end{pmatrix}$ if and only if $Bv - B^t v = i \mu v$, so $\mu = 2 \operatorname{Im} (\eta)$, where $\eta$ is an eigenvalue of $B$ and $v$ is an eigenvector of $B-B^t$.
So, if you choose more than $n$ real numbers, or choose any complex numbers $\lambda$ that do not satisfy $\bar{\lambda} = \pm \lambda$, or you choose more than $n$ purely imaginary numbers, then they cannot be eigenvalues of $A$.
Now, in the odd dimensional case $n=2m+1$, $A$ takes the form
$$A= \begin{pmatrix} B & u & B^t P \\ u^t & a & (Pu)^t \\
PB^t& Pu & PBP \end{pmatrix}$$
for some matrix $B$, vector $u$ and real number $a$. For $\lambda$ a real eigenvalue of $A$, we have the eigenvector $\begin{pmatrix} v \\ b \\Pv \end{pmatrix}$, so
$$(B+B^t)v + bu = \lambda v$$
and
$$2(u,v)+ab=\lambda b.$$
Choosing $u=0$ gives
$$(B+B^t)v=\lambda v$$
and
$ab=\lambda b.$
So we can pick $m$ eigenvectors for $B+B^t$ as above, and also the vector given by $v=0$ and $b=1$, which has eigenvalue $a$. The imaginary eigenvalues $i \mu$ have eigenvectors $\begin{pmatrix} v\\0\\-Pv \end{pmatrix}$, and so
$$(B-B^t)v=i\mu v,$$
with the other equation being $0=0$. So as before $\mu$ is twice the imaginary part of an eigenvalue of $B$.
