Matrices such that $\det(p(A)-p(0))=p(\det A)-p(0)$ for all polynomial $p$ 
Question: Find all $2\times 2$ matrix $A$ such that
  $$\det(p(A)-p(0))=p(\det A)-p(0)$$
  for all polynomial $p$.

The zero matrix works since both sides are obviously zero. But I cannot find any other example. This seems like a very strong condition to met, so my guess is that $0$ is the only solution, but I coulnd't find an argument to support this intuition.
What I got so far is this:
Taking $p(x)=2x^n$ for $n\ge 1$, we get
$$4\det(A^n)=\det(2A^n)=2\det(A^n)\implies\det(A^n)=0.$$
So $A^n$ is singular for all $n\ge 1$. 
 A: $A^n$ is singular for all $n\geq 1$ if and only if $A$ itself is singular. You have proven nicely that $A$ needs to be singular, so you can now safely assume that $A$ is a rank $1$ matrix (or rank $0$, but that's boring). 
We have two options. First off, if $A$ is diagonalizable:
OK, so $A=PLP^{-1}$ where $L=\begin{bmatrix}\lambda & 0\\ 0 & 0\end{bmatrix}$ for some number $\lambda$. That means that $$p(A) = Pp(L) P^{-1}$$ and that
$$p(A) - p(0) = P(p(L) - p(0))P^{-1}$$
so $\det(p(A) - p(0)) = \det(p(L) - p(0))$
and this number must be equal to $p(\det(A)) - p(0) = p(0) - p(0)=0$
for all polynomials $p$.
Now, you can probably show that if $\lambda \neq 0$, then there exists some polynomial $p$ that $\det(p(L) - p(0))$ will not be equal to $0$, so you can conclude that $\lambda$ must be $0$ and that $0$ is the only diagonalizable matrix you can find.

If $A$ is not diagonalizable, then both its eigenvalues are equal to $0$ and Jordan's canonical form will tell you that $A=PLP^{-1}$ where $L = \begin{bmatrix}0&1\\0&0\end{bmatrix}$
Now let's say $p(x) = a_0 +a_1 x + a_2x^2 +\dots + a_nx^n$. Since $L^2=0$, you get $$p(L)-p(0) = a_1 L + a_0I - a_0I= a_1 L$$ and $\det(p(L) - p(0)) = \det(a_1L) = 0$ for all polynomials $p$. So in this case, any invertible matrix $P$ gives you one possible candidate for $A$.
A: An $n \times n$ matrix $A$ (for $n \geq 2$) will satisfy 
$$
\det(p(A) - p(0)) = p(\det(A)) - p(0)
$$
for all polynomials $p$ if and only if it is singular.
To show that any such $A$ is singular, it suffices to consider the polynomial $p(x) = 2x$ (as you have done).  We then need to show that the condition holds for any singular matrix.
So, suppose that $A$ is singular, which is to say that $\det(A) = 0$.  From there, we note that for any polynomial $p$, the polynomial $p(x) - p(0)$ can be factored into the polynomial 
$$
p(x) - p(0) = x\,q(x)
$$
for some polynomial $q(x)$. We may then rewrite the desired equality as
$$
\det(A q(A)) = \det(A)q(\det(A))
$$
Note, however, that the left hand side can be rewritten as $\det(A)\det(q(A))$.  So, if $\det(A) = 0$, then we have equality, regardless of which polynomial $q$ happens to be.
The conclusion follows.
