Can we deduce the characteristic polynomial for this matrix?

The answer is no.

You are given a degree $n$ polynomial $p(x)$ that an $n\times n$ matrix $A$ satisfies. This is not enough information to find the characteristic polynomial $c_A(x)$, although you will be able to narrow it down to finitely many possibilities.

Let's look at an example to see why. Suppose your friend picks the matrix $A$. Suppose he doesn't tell you $A$, but he does tell you $p(x)$ and asks you to guess $c_A(x)$. Suppose your friend picked

$$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$

and told you that $p(x)=x^3-6x^2+11x-6$.

You could then reason that $p(x)=(x-1)(x-2)(x-3)$. Which would mean that the minimal polynomial $m_A(x)$ must be one of the following: $(x-1)$, $(x-2)$, $(x-3)$, $(x-1)(x-2)$, $(x-1)(x-3)$, $(x-2)(x-3)$, $(x-1)(x-2)(x-3)$.

Hence the characteristic polynomial $c_A(x)$ must be one of the following: $(x-1)^3$, $(x-2)^3$, $(x-3)^3$, $(x-1)^2(x-2)$, $(x-1)(x-2)^2$, $(x-1)^2(x-3)$, $(x-1)(x-3)^2$, $(x-2)^2(x-3)$, $(x-2)(x-3)^2$, $(x-1)(x-2)(x-3)$.

Since your friend picked

$$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$

the characteristic polynomial is $c_A(x)=(x-1)^2(x-2)$, but you can't prove that, because for all you know he may have picked

$$B=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$

which also satisfies $p(x)=x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$.

Take matrix $$ A = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{array} \right) $$ Then $$ A^4 - 11 A^3 + 41 A^2 - 61 A + 30 I = 0$$ However, the characteristic polynomial is $$ x^4 - 6x^3 + 13x^2 - 12x + 4 $$ and the minimal polynomial is $$ x^2 - 3 x + 2 $$

The zero matrix satisfies every homogeneous polynomial, but clearly not every homogeneous polynomial is the characteristic polynomial of the zero matrix.