Is it possible to turn an eulerian NFA into a linear size DFA?
I don't think so.

Suppose you could. Then you can also do this:


Take any NFA $\mathcal A$ with alphabet $\Sigma$, states $Q$, transitions $\Delta$, initial states $I$ and final / acception states $F$.
Add transitions to make the graph eulerian in the following way. Take any enumeration of the transition $p_i\overset{a_i}{\to}q_i$. For each $i$, add a transition $q_i\overset{b_i}{\to}p_{i+1}$ where $b_i$ is a new letter (and the indices are modulo the number of transitions). Note that you add $|\Delta|$ letters and transitions so this transformation is linear.
Add transitions to make the graph strongly connected in the following way. Take any enumeration of the states $s_j$. Add a new letter $c$. For each $j$, add a transition $s_i\overset{c}{\to}s_{i+1}$. In other words, add a Hamiltonian cycle using only the new letter $c$. Note that this transformation adds $1$ letter and $|Q|$ transitions so it's linear.
Use your magic procedure to build a new deterministic automaton $\mathcal B'$ so that it recognizes the same language as $\mathcal A'$ (the modified automaton) and $|\mathcal B'|=O(|\mathcal A|)$.
Remove added letters ($b_i$ for each $i$ and $c$) and all transitions using them. Call the new automaton $\mathcal B$.


Since $\mathcal L (\mathcal A) = \mathcal L (\mathcal A')\cap \Sigma^* $ (because if you don't use the new letters, then you're taking a path that already existed), and similarly $\mathcal L (\mathcal B) = \mathcal L (\mathcal B')\cap \Sigma^* $, we have $\mathcal L (\mathcal A) = \mathcal L (\mathcal A')\cap \Sigma^*=\mathcal L (\mathcal B')\cap \Sigma^*=\mathcal L (\mathcal B)$ with $\mathcal B$ a DFA so that $|\mathcal B|=O(|\mathcal A|)$.

But this can not be because of the usual counterexample where the size of the corresponding DFA is of exponential size. To get an explicit counterexample, you can apply the procedure to that counterexample.