If you define the characteristic polynomial of a matrix $A$ to be $\det(xI - A)$, then for $M$ invertible we have:
$\det(xI - M^{-1} A M)$
$= \det(M^{-1} xI M - M^{-1} A M)$
$= \det(M^{-1} (xI-A) M)$
$= \det (M^{-1}) \det(xI-A) \det(M)$
$=\det(xI - A)$
The answer is no in general since Hermitian matrices have real eigenvalues. So for example, there is no Hermitian matrix whose characteristic polynomial is $X^2+1$.
Even if you restrict to polynomials with real roots, I doubt you can find a simple formula : I think that if there is a rational formula that works for every polynomial with real roots, it should also apply to all polynomials. Let me explain what I mean precisely. We'll use the following lemma which states that when rational equations hold for polynomial with real roots, they should hold everywhere. Denote $\mathbb{C}_{n,u}[X]$ the set of monic complex polynomials of degree $n$.
Lemma : Let $f$ be a rational function
$$\begin{eqnarray*}f :& \mathbb{C}_{n,u} & \longrightarrow \mathbb{C}^N\\
& P &\longmapsto f(P)\end{eqnarray*}$$
such that $f$ is zero on polynomial with real roots. Then $f$ is zero.
Proof :
The map
$$\begin{eqnarray*}p_n :& \mathbb{C}^n & \longrightarrow \mathbb{C}_{n,u}[X]\\
& (\lambda_1, \ldots, \lambda_n) &\longmapsto p_n(\lambda_1, \ldots, \lambda_n) = \prod_{i = 1}^n (X-\lambda_i)\end{eqnarray*}$$
is also a rational map (meaning the coefficients of a polynomial are rational functions of its roots, and we can actually compute them, those are called elementary symmetric polynomials). By hypothesis, the function $f \circ p_n$ is zero on $\mathbb{R}^n$. But since $f \circ p_n$ is rational, it is actually zero everywhere, and because $p_n$ is surjective (every polynomial is split in $\mathbb{C}$), then $f$ is zero.
Now assume there is a rational function
$$\begin{eqnarray*}A_n :& \mathbb{C}_{n,u}[X] & \longrightarrow \mathcal{M}_n(\mathbb{C})\\
& P &\longmapsto A_n(P)\end{eqnarray*}$$
(by that I mean that the coefficients of the matrix $A_n(P)$ are rational functions of the coefficients of $P$) such that the characteristic polynomial of $A_n(P)$ is $P$ whenever $P$ has real roots. This means the rational function $P \longmapsto \det(XI_n - A_n(P)) - P$ is zero on polynomial with real roots, so it's zero everywhere, which means the characteristic polynomial of $A_n(P)$ is always $P$ without assumption on the roots of $P$.
Assume moreover that $A_n(P)$ is hermitian whenever $P$ has real roots. Then I claim $A_n(P)$ is hermitian for any real polynomial $P$ from the same kind of argument. Indeed denote $A_n^* : P \longmapsto \left[A_n(P^*)\right]^*$, where $P^*$ denotes the complex conjugate of the polynomial $P$, and $\left[A_n(P^*)\right]^*$ is the conjugate transpose of the matrix $A_n(P^*)$. This is a rational function (beware that $P \mapsto [A_n(P)]^*$ is not rational however !). Our assumption is that the rational function $A_n - A_n^*$ is zero on all polynomial with real roots, so it's zero everywhere. When $P$ is real, this means $A_n(P)$ is hermitian (because $A_n(P) = A_n^*(P) = [A_n(P)]^*$, the last equality being only true when $P$ is real). This yields a contradiction if $P$ is a real polynomial with complex roots.
Best Answer
(For sake of having an answer, let's turn the comments to this question into one.)
The minimal polynomial and characteristic polynomial of a companion matrix are equal. So, if $A\sim A_c$, the minimal and characteristic polynomials of $A$ are equal, too. In other words, in the Jordan form of $A$, every eigenvalue is associated with only one Jordan block. Alternatively speaking, the geometric multiplicity of each eigenvalue is equal to $1$.
When $A$ is Hermitian, geometric multiplicities and algebraic multiplicities coincide. Hence the above necessary and sufficient condition reduces to that all eigenvalues of $A$ are simple, i.e. $A$ has distinct eigenvalues.