The Birman–Hilden Theorem and the Nielsen–Thurston classification

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So this post is half question/half reference request, as I'm sure it's the kind of thing people would have thought about before (and indeed the question might even be trivial), but I've been unable to find anything on it.

Recall that the Birman–Hilden theorem (cf. e.g. Section 9.4 of Farb and Margalit's wonderful book) defines for us a surjective map $\varphi: \mathrm{SMod}(S_{g,0}) \to \mathrm{Mod}(S_{0,2g+2})$ with kernel $\langle \iota \rangle$. Here, $S_{g,n}$ denotes a surface of genus $g$ with $n$ punctures, $\mathrm{Mod}(S_{0,2g+2})$ the mapping class group of $S_{0,2g+2}$, and $\mathrm{SMod}(S_{g,0})$ the so-called symmetric mapping class group consisting of orientation preserving homeomorphisms (up to isotopy) of the surface that commute with a particular order two homeomorphism $\iota$.

Now, my question is the following: Has anyone written down (or considered) what happens to the dictionary of the Nielsen–Thurston classification under the Birman–Hilden map $\varphi$? It is clear that finite-order mapping classes go to finite-order ones, for reducible mapping classes one probably needs to think a little bit more, and the same goes for pseudo-Anosovs.

Edit: (Very) partially answering my own question, in the case $g = 2$, where we also have $\mathrm{SMod}(\Sigma_{2,0}) = \mathrm{Mod}(\Sigma_{2,0})$, Whittlesey has shown that reducible mapping classes are mapped to reducible mapping classes on the sphere. The proof uses the Nielsen–Thurston classification and as such does not say much about what happens to pseudo-Anosovs, although it defines explicitly a nice class of pseudo-Anosovs in both $\mathrm{Mod}(\Sigma_{2,0})$ and $\mathrm{Mod}(\Sigma_{0,6})$. One would expect that for at least some general class of pseudo-Anosovs in $\mathrm{Mod}(\Sigma_{2,0})$ to project to pseudo-Anosovs (with the same/directly related dilatation).

Edit #2: Those interested should also check out this Reddit crosspost.