Why are $M$-estimators NOT scale equivariant? Consider the following location model.
$$
x_i = \mu + u_i, (i = 1,\dots, n),
$$
where $u_i$ are $i.i.d.$ with density function $f_0$. Hence, $x_i$ are $i.i.d.$ with density function $f_0(x-\mu)$. It usually is of interest to find the estimator of $\mu$. One example is the Maximum-Likelihood-Estimator for location $\mu$ (a special case of $M$-estimators) defined to be 
$$
\widehat\mu(x_1, \dots, x_n) := \arg\min_\mu \sum_{i=1}^n -\log f_0(x_i-\mu).
$$
I read that this estimator is NOT scale equivariant. That is, for a constant $c \in \mathbb R$,
$$
\widehat\mu(cx_1, \dots, cx_n) \neq c\widehat\mu(x_1, \dots, x_n). (*)
$$
I do not understand why this is the case. I suppose whether $(*)$ holds depends on the function $f_0$, doesn't it? Could anyone explain this to me, please? Some examples are appreciated. Thank you!
UPDATE: As an example, let us consider the normal distribution $N(\mu, \sigma^2)$. The MLE for $\mu$ (regardless of $\sigma$) can be shown to be
$$
\widehat\mu(x_1, \dots, x_n) = \frac{\sum_{i=1}^n x_i}{n}.
$$
Moreover, if you multiply $x_i$ by $c$, it is also straightforward to show that
$$
\widehat\mu(cx_1, \dots, cx_n) = \frac{c\sum_{i=1}^n x_i}{n} = c\widehat\mu(x_1, \dots, x_n).
$$
Also note in all these derivation one did not use $\sigma$ at all. Hence, I do NOT see why this estimator is not scale equivariant. Could anyone give me an example where the identity 
$$
\widehat\mu(cx_1, \dots, cx_n) = c\widehat\mu(x_1, \dots, x_n).
$$
does NOT hold, please? Thank you!
 A: Because you need it to be a location and scale family, $f((x - \mu)/\sigma)$, and the estimation method needs to include a data-based estimator of $\sigma$ as well, e.g., the MAD (median absolute deviation of the residuals). For most but the normal case, the scale parameter makes a difference in the log of the PDF.
Addendum
Now that I'm not distracted by the NCAA basketball finals, I can add some info, hopefully with fewer mistakes...
The next step (for sufficiently smooth densities) in the question's second displayed equation (with $\arg\min$) is to differentiate and set to zero -- i.e., solve the equation
$$ \sum \psi(x_i - \hat\mu) = 0 $$
where $\psi(z) = -\frac d{dz}\log f_0(z)$. If $f_0(z) = \frac1{\sqrt{2\pi}\sigma}\exp\{-\frac12z^2/\sigma^2\}$, we have $-\log f_0(z) = \frac12\log(2\pi)  + \log\sigma + \frac12z^2/\sigma^2$, so that $\psi(z) = z/\sigma^2$ -- and thus we need to solve the equation $\frac1{\sigma^2}\sum(x_i-\hat\mu) = 0$. Since $\sigma\ne0$, we can just canel it out and it plays no further role in solving the equation. Thus, $\hat\mu$ is scale equivariant.
However, suppose that $f_0(z) = \frac1{\pi\sigma}\cdot\frac1{1 + z^2/\sigma^2}$ is the Cauchy density with $\sigma$ as a scale parameter. We now have $-\log f_0(z) = \log{\pi\sigma} + \log(1+z^2/\sigma^2)$, so that $\psi(z) = \frac{2z/\sigma^2}{1+z^2/\sigma^2} = \frac{2z}{\sigma^2+z^2}$. In solving $\sum\psi(x_i-\hat\mu)=0$, $\sigma$ does not cancel out, and its value affects the solution! That is, with this Cauchy density as the basis for the M estimation, $\hat\mu$ is not scale equivariant.
The gist of my original answer is that you should include the scale parameter in the equation:
$$ \sum \psi\{(x_i - \hat\mu)/\hat\sigma\} = 0 $$
The classic M-estimation example uses Huber's $\psi(z) = \mbox{sign}(z)\cdot\min(|z|,c)$ where $c$ is a suitable tuning constant, e.g. 1.34. A popular scale estimator is $\hat\sigma = \mbox{med}|x_i - \mbox{med}_j\{x_j\}|$, then obtain $\hat\mu$ as an iteratively weighted mean with weights $w_i = \frac{\psi\{(x_i - \hat\mu)/\hat\sigma\}}{(x_i - \hat\mu)/\hat\sigma}$ (and with, say, the median as the initial $\hat\mu$. This estimator is scale equivariant because $\hat\sigma$ is scaled proportionally to the scale of the data.
