Consider
$$z \mapsto \frac{(n-2) z^n + n z}{n z^{n-1} + (n-2)}.$$
This has $n+1$ fixed points, at $0$, $\infty$, and the $(n-1)$-st roots of $-1$. The only critical points are the roots of $-1$, each of which is ramified of index $3$. So this is a map with all critical points fixed, and all fixed points but two critical.

I am tempted to leave it at that. But, being a nice person, I will explain how I found this. Moreover, I will show that this is (up to conjugacy), the only degree $n$ map with $n+1$ distinct fixed points, two of which are not critical and the rest of which are critical with multiplicity $2$ (ramification index $3$.)

We can take the two noncritical fixed points to be $0$ and $\infty$. So our map is of the form
$$z \mapsto z - z p/q,$$
where $p$ and $q$ are of degree $n-1$ and relatively prime. The $n-1$ fixed points other than $0$ and $\infty$ are the roots of $p$.

The derivative of this map is
$$\frac{q^2 - zqp' + zpq' - pq}{q^2}.$$
The condition that the fixed points other than $0$ and $\infty$ be critical means $p$ divides the numerator. So $p$ divides $q (q-zp')$ and, as $p$ and $q$ are relatively prime, we conclude that $q - z p' = kp$. Checking degrees, $k$ has degree $0$ and is thus a constant, to be determined later.

Now, we want to impose the stronger condition that every zero of $p$ be doubly a critical point, so the numerator is $\ell p^2$ for some constant $\ell$.
Plugging in $q = kp + z p'$, and simplifying
$$\ell p^2 = p \left( k(k-1) p + 2 k z p' + z^2 p'' \right).$$
Cancelling $p$ from both sides,
$$\ell p = k(k-1) p + 2 k z p' + z^2 p''.$$
Plugging in $z=0$, and noting that $p(0) \neq 0$, we get $\ell = k(k-1)$. So
$$2 kz p' + z^2 p'' =0.$$
The solution to this differential equation is $p = C z^{1-2k} + D$. But we know that $z$ has degree $n-1$, so $1-2k=n-1$ and $k = -(n-2)/2$.

Taking the simplest choices $C=D=1$ and plugging back in gives the above solution. All other solutions are related to this one by rescaling the variable $z$.

I was thinking about this very old question again, due to conversations with Sarah Koch. Here is another infinite family of examples which we can write down explicitly: Let $g(x)$ be the degree $n$ Gegenbauer polynomial satisfying
$$g'' = \frac{n(n-1)}{x^2-1} g.$$
Let $f(x)$ be the Newton recursion
$$f(x) = x-g/g'.$$
Then
$$f' = \frac{g g''}{(g')^2} = \frac{n (n-1) g^2}{(x^2-1) (g')^2} .$$
So every critical point of $f$ is a root of $g$, and hence a fixed point of $f$.

More specifically, $\pm 1$ are critical points of multiplicity $1$ and the other $n-2$ roots of $g$ have multiplicity $2$. The only fixed point which is not critical is the fixed point at $\infty$.

On reading further, I see that this example is in section 11 of the Cordwell et al paper.

There has been a lot of research on this question in the last few years. As well as the paper of Cordwell et al linked by the OP, see the Ph. D thesis of Hlushchanka and this paper of his.