Theorem. Let $T$ be an operator on the finite dimensional complex vector space $\mathbf{W}$. The characteristic polynomial of $T$ equals the minimal polynomial of $T$ if and only if the dimension of each eigenspace of $T$ is $1$.
Proof. Let the characteristic and minimal polynomial be, respectively, $\chi(t)$ and $\mu(t)$, with
$$\begin{align*}
\chi(t) &= (t-\lambda_1)^{a_1}\cdots (t-\lambda_k)^{a_k}\\
\mu(t) &= (t-\lambda_1)^{b_1}\cdots (t-\lambda_k)^{b_k},
\end{align*}$$
where $1\leq b_i\leq a_i$ for each $i$. Then $b_i$ is the size of the largest Jordan block associated to $\lambda_i$ in the Jordan canonical form of $T$, and the sum of the sizes of the Jordan blocks associated to $\lambda_i$ is equal to $a_i$. Hence, $b_i=a_i$ if and only if $T$ has a unique Jordan block associated to $\lambda_i$. Since the dimension of $E_{\lambda_i}$ is equal to the number of Jordan blocks associated to $\lambda_i$ in the Jordan canonical form of $T$, it follows that $b_i=a_i$ if and only if $\dim(E_{\lambda_i})=1$. QED
In particular, if the matrix has $n$ distinct eigenvalues, then each eigenvalue has a one-dimensional eigenspace.
Also in particular,
Corollary. Let $T$ be a diagonalizable operator on a finite dimensional vector space $\mathbf{W}$. The characteristic polynomial of $T$ equals the minimal polynomial of $T$ if and only if the number of distinct eigenvalues of $T$ is $\dim(\mathbf{W})$.
Using the Rational Canonical Form instead, we obtain:
Theorem. Let $W$ be a finite dimensional vector space over the field $\mathbf{F}$, and $T$ an operator on $W$. Let $\chi(t)$ be the characteristic polynomial of $T$, and assume that the factorization of $\chi(t)$ into irreducibles over $\mathbf{F}$ is
$$\chi(t) = \phi_1(t)^{a_1}\cdots \phi_k(t)^{a_k}.$$
Then the minimal polynomial of $T$ equals the characteristic polynomial of $T$ if and only if $\dim(\mathrm{ker}(\phi_i(T)) = \deg(\phi_i(t))$ for $i=1,\ldots,k$.
Proof. Proceed as above, using the Rational Canonical forms instead. The exponent $b_i$ of $\phi_i(t)$ in the minimal polynomial gives the largest power of $\phi_i(t)$ that has a companion block in the Rational canonical form, and $\frac{1}{d_i}\dim(\mathrm{ker}(\phi_i(T)))$ (where $d_i=\deg(\phi_i)$) is the number of companion blocks. QED
Before proving $AB$ and $BA$ have the same characteristic polynomials show that if $A_{m\times n}$ and $B_{n\times m} $ then characteristic polynomials of $AB$ and $BA$ satisfy following statement: $$x^n|xI_m-AB|=x^m|xI_n-BA|$$ therefore easily conclude if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.
Define $$C = \begin{bmatrix} xI_m & A \\B & I_n \end{bmatrix},\ D = \begin{bmatrix} I_m & 0 \\-B & xI_n \end{bmatrix}.$$ We have
$$
\begin{align*}
\det CD &= x^n|xI_m-AB|,\\
\det DC &= x^m|xI_n-BA|.
\end{align*}
$$
and we know $\det CD=\det DC$ if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.
Best Answer
Let $k$ be an algebraically closed field.
Facts: Let $M$ be a matrix with coefficients in $k$. Then $M$ is conjugate to a block-diagonal matrix where each block is a Jordan block. These blocks, up to reordering, classify the conjugacy classes. Moreover, let $\lambda$ be an eigenvalue of $M$. Then
Therefore, it is enough to look for the smallest integer $N$ such that there are an integer $k$, and two different $k$-tuples of integers $(n_1,\cdots,n_k)$ and $(m_1,\cdots,m_k)$ such that:
and to form two matrices with only one eigenvalue having Jordan blocks of sizes $(n_1,\cdots,n_k)$ and $(m_1,\cdots,m_k)$.
It is easily seen that the smallest $N$ is $7$, and that examples of tuples are $(3,3,1)$ and $(3,2,2)$.
An example of a couple of two matrices is the following:
$ \begin{pmatrix} 1&1&0&0&0&0&0\\ 0&1&1&0&0&0&0\\ 0&0&1&0&0&0&0\\ 0&0&0&1&1&0&0\\ 0&0&0&0&1&1&0\\ 0&0&0&0&0&1&0\\ 0&0&0&0&0&0&1\\ \end{pmatrix}$ and
$\begin{pmatrix} 1&1&0&0&0&0&0\\ 0&1&1&0&0&0&0\\ 0&0&1&0&0&0&0\\ 0&0&0&1&1&0&0\\ 0&0&0&0&1&0&0\\ 0&0&0&0&0&1&1\\ 0&0&0&0&0&0&1\\ \end{pmatrix}.$