I assume our OP Avenger means "every non-zero vector of ${\mathbb{C}}^n$ is an eigenvector of A" instead of
"every non-zero vector of ${\mathbb{C}}^n$ is an eigenvalue of A"; otherwise, the problem makes no sense.
If every non-zero vector $w$ is an eigenvector, each with eigenvalue $\mu_w$,
$Aw = \mu_w w \tag 1$
by choosing two linearly independent vectors $u$ and $v$ we have
$\mu_u u + \mu_v v = Au + Av = A(u + v)$
$= \mu_{u + v}(u + v) = \mu_{u + v}u + \mu_{u + v}v; \tag 2$
linear independence of $u$ and $v$ then forces
$\mu_u = \mu_{u + v} = \mu_v; \tag 3$
thus all eigenvalues are equal to some
$\mu \in \Bbb C, \tag 4$
that is, for any vector $w$,
$Aw = \mu w, \tag 5$
or
$A = \mu I; \tag 6$
$A$ is a scalar multiple of $I$.
The characteristic polynomial of $A$ is
$\det(A - xI) = \det (\mu I - xI)$
$= \det((\mu - x)I) = (\mu - x)^n; \tag 7$
however, $A = \mu I$ also satisfies its minimal polynomial
$m(x) = \mu - x, \tag 8$
and
$m(x) \ne (\mu - x)^n \tag 9$
unless $n = 1$.
So we have items (1) and (3) in the OP's question are true, but (2) and (4) are false.
When $T(U)=U$ for a $3$-dimensional subspace $U\subseteq\mathbb R^n$, the restriction $T|_U\colon U\to U$ has $3$ of the eigenvalues of $T$ (counting algebraic multiplicity). The non-real complex eigenvalues of a real linear transform always come in pairs: When $\lambda\in\mathbb C$ is a complex eigenvalue, then so is the complex conjugate $\overline \lambda$. Hence, the number of non-real complex eigenvalues of any real linear transform is even. Since $3$ is odd, $T|_U$ has one or three real eigenvalues.
Best Answer
HINT: It’s easy to see that if $A=\lambda I$, then every non-zero vector in $\Bbb C^n$ is an eigenvector of $A$ for the eigenvalue $\lambda$, so it’s certainly possible for all of the eigenvalues of $A$ to be equal; this shows you that (2) and (4) are not necessarily true, while (1) and (3) can be true. All that remains is to decide whether (3) must be true (in which case (1) will also be true).
Suppose that $x,y\in\Bbb C^n$, $Ax=\lambda_1x$, and $Ay=\lambda_2y$, where $\lambda_1\ne\lambda_2$. Then
$$A(x+y)=Ax+Ay=\lambda_1x+\lambda_2y\;,$$
and $A(x+y)=\lambda(x+y)$ for some $\lambda\in\Bbb C$, so $\lambda x+\lambda y=\lambda_1x+\lambda_2y$, and therefore
$$(\lambda-\lambda_1)x=(\lambda_2-\lambda)y\;.$$
Observe that at least one of $\lambda-\lambda_1$ and $\lambda_2-\lambda$ is non-zero and derive a contradiction, thereby showing that all eigenvalues of $A$ must be equal and hence that (1) must be true.
Now if all of the eigenvalues are equal to $\lambda$, say, so that $Ax=\lambda x$ for each $x\in\Bbb C^n$, then $A-\lambda I=0$ for all $x\in\Bbb C^n$, so what must $A$ be?