The question

Let $f$ be a nonconstant polynomial over $\mathbb{C}$.  Let's say that a point $c \in \mathbb{C}$ is unusual for $f$ if every root $x$ of $f(x) - c$ is repeated.  Can $f$ have more than one unusual point?

Short remarks


There can be exactly one unusual point, e.g. if $f(x) = x^2$.  There can be none, e.g. if $f(x) = x^3 + 3x$.
There are at most $\deg(f) - 1$ unusual points, since every unusual point is the image under $f$ of a critical point. 
The hypotheses "nonconstant" and "over $\mathbb{C}$" could be varied.  I added them to rule out cases like the following: over a field $k$ of characteristic $p$, every point of $k$ is unusual for $x^p$ (since the derivative of $x^p$ is $0$).  I'd be happy to change the hypotheses to "nonconstant polynomial over an algebraically closed field of characteristic 0".  Maybe something like "polynomial whose derivative is nonzero, over an algebraically closed field" would also be sensible.


Weak reason to expect the answer to be "no"

Perhaps there's a very short answer to my question: I could be overlooking something elementary.  But in case it's not so easy, I'll give a flimsy argument for why we might expect the answer to be "no" - that is, for why we might expect every polynomial to have at most one unusual point.  

My question is equivalent to: is there a nonconstant polynomial $f$ over $\mathbb{C}$ for which $1$ and $-1$ are both unusual?  For $1$ to be unusual means that every root of $f(x) - 1$ is also a root of $f'(x)$.  Writing $d = \deg(f)$, this is equivalent to 
$$
(f(x) - 1) \mid f'(x)^d.
$$
Similarly, for $-1$ to be unusual means that $(f(x) + 1) \mid f'(x)^d$.  Both together are equivalent to 
$$
(f(x)^2 - 1) \mid f'(x)^d,
$$
that is,
$$
(f(x)^2 - 1)\cdot g(x) = f'(x)^d
$$
for some $g(x) \in \mathbb{C}[x]$.  This forces $\deg(g) = d(d - 3)$ (and so $d \geq 3$).  

So, can we find $f$ and $g$ satisfying the last displayed equation?  Comparing coefficients, what we have here is a system of $d(d - 1) = d^2 - d$ equations in $(\deg(f) + 1) + (\deg(g) + 1) =  d^2 - 2d + 2$ unknowns.  There are $d - 2$ more equations than unknowns, and $d \geq 3$, so a first guess is that it can't be done. 

This is impossible by the Mason-Stothers theorem (which holds over any algebraically closed field of characteristic zero). 

We want to find $f, g, h$ such that $f + g = h$ where $g$ is a constant and $f, h$ have all of their roots repeated. If $g$ is nonzero, $f, h$ must be relatively prime. Letting $d = \deg f$, it follows that $fgh$ has at most $d$ roots, but by Mason-Stothers $fgh$ must have at least $d+1$ roots; contradiction. 
