Veech group and mapping class group

So I have this question which is somewhat directed to people knowing a bit about translation surfaces. I am sure it is only a technical issue.

I consider $f : (\Sigma, \omega) \longrightarrow (\Sigma, \omega)$ an affine automorphism of $(\Sigma, \omega)$, that is who writes down as an element of $\mathrm{SL}(2, \mathbb{R}) \ltimes \mathbb{R}^2$ in coordinates charts. I assume that $f$ is not the identity. Is $f$ necessarily non-trivial in $\mathrm{MCG}(\Sigma)$ the mapping class group of $\Sigma$?

The question is actually can $f$ be isotopic to the identity? The natural way to tackle the question seems to me to look at the action of $f$ on the saddle connections. This way one easily proves that if $f$ (or a suitable power that fixes the singular points) is isotopic to the identity relatively to the set of singular points then it must be the identity (because the image of any saddle connection must be itself).

The answer to your question is "yes, such a $f$ is non-trivial in $MCG(\Sigma)$".
Indeed, if the linear part of $f$ is non-trivial, this is Proposition 2.5 in Veech's paper (http://www.ams.org/mathscinet-getitem?mr=1005006).
If the linear part of $f$ is trivial (i.e., $f$ is an automorphism of a translation surface), this is a consequence of Lefschetz formula (as briefly explained at page 456 of my paper [http://w3.impa.br/~cmateus/files/matheus-yoccoz1.pdf] with Yoccoz, for instance) when the genus of $\Sigma$ is $>1$.