Question: The Cayley–Hamilton theorem states that every square matrix satisfies its own characteristic equation.

But does it work in the opposite direction?

If for example for a certain matrix $A$ we know that

$ A^2-6A+9I=0, $

does that mean that the characteristic equation of $A$ is

$
\lambda^2-6\lambda+9=0
$ 
?

Answer: 
Even without counterexamples it is obvious that your statement can't be true because if $A$ is a root of the polynomial $p(x)$ then it must be the root of $p(x)q(x)$ for any polynomial $q$. So that way we would get the matrix $A$ has infinitely many characteristic polynomials.
No. In general if a matrix is a root of a polynomial, that polynomial is a multiple of the minimal polynomial of that matrix. 

Even if you only consider n-degree polynomials for an n×n matrix, this still can fail. 

For example, if A=0, then the characteristic polynomial is $x^n$, but A is the root of any polynomial $xg(x)$.

However, all is not lost. If your minimal polynomial equals your characteristic polynomial (as with companion matrices or in the case you have no repeated roots), then the characteristic polynomial (= minimal polynomial) will be the unique monic n-degree polynomial with root A.