The statement is poorly phrased, but the idea that I think was intended is correct. In particular, we have the following:
Claim: A matrix $M$ has equal characteristic and minimal polynomials if and only if for each of its eigenvalues, $M$ has only one Jordan block.
For more information on matrices like these, see Horn and Johnson's Matrix Analysis. In that context, such matrices are referred to as "non-derogatory".
To see that the statement holds, it suffices to understand how the minimal and characteristic polynomials relate to the Jordan form of $M$. In particular, suppose that the minimal polynomial of $M$ is given by
$$
p(x) = (x-\lambda_1)^{m_1} \cdots (x - \lambda_k)^{m_k}
$$
where $\lambda_1,\dots,\lambda_m$ are distinct. For each $j = 1,\dots,k,$ $m_j$ is the size of the largest Jordan block associated with $\lambda_j$.
On the other hand, the characteristic polynomial is given by
$$
\chi(x) = (x - \lambda_1)^{d_1} \cdots (x - \lambda_k)^{d_k}.
$$
In general, because $p(x) \mid \chi(x)$, it must be that $d_j \geq m_j$ for all $j = 1,\dots,k$. Note that for each $j$, $d_j$ is the sum of the sizes of all Jordan blocks associated with $\lambda_j$.
With these characterizations, it is clear that if $\lambda_j$ has more than one Jordan block in the Jordan form, then it must hold that $d_j > m_j$. Equivalently, if $d_j = m_j$ for all $j$ (so that $p = \xi$), then each $\lambda_j$ has only one Jordan block in the Jordan form.
Best Answer
The roots of the minimal polynomial are precisely the eigenvalues, and the degree of a nonzero polynomial is at least the number of its roots, so 1. is true.
With your conditions, it is possible for the minimal polynomial to have any degree between 4 (in the diagonalisable case) and 10, so all the other answers are false.
To prove this, let's call $J_k = \begin{pmatrix} 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 \end{pmatrix} \in M_k(\mathbb C)$ the usual nilpotent Jordan block.
You can easily check that the minimal polynomial of $J_k$ is $X^k$.
For $1 \leq k \leq 7$, the block-diagonal matrix $\mathrm{diag}(1,2,3) \oplus J_k \oplus \mathrm{diag}(\underbrace{0,\ldots,0}_{\text{$7-k$ zeroes}}) \in M_{10}(\mathbb C)$ has four eigenvalues, and its minimal polynomial is $X^k(X-1)(X-2)(X-3)$.