Let $X$ be a fixed parametrizing space. Let $E$ and $E'$ be two spectra and let $E\_X$ and $E'\_X$ be their trivial parametrized versions. Intuitively I imagine that the morphisms of parametrized spectra $E\_X\to E'\_X$ should correspond to maps $X\to \operatorname{Map}(E, E')$ where the mapping space on the right should be the underlining infinite loop space of the function spectrum, if you wish $\operatorname{Map}(E, E')=\Omega^\infty F(E, E').$


Is this, or at least anything similar to it, actually true? 



Are morphisms of parametrized spectra themselves parametrized morphisms of spectra?
This is true.


To prove it I will use the fact that maps of parametrized spectra can be computed as natural transformations of functors from $X$ into the $\infty$-category of spectra (cfr. [this paper](https://arxiv.org/abs/1403.4325)) and the formula for computing the space of natural transformation (e.g. see [here](https://arxiv.org/abs/1408.3065), proposition 2.3):


$$\mathrm{Map}(E\_X,E'\_X) = \int\_{x\in X} \mathrm{Map}(E\_x,E'\_x) = \int\_{x\in X} \mathrm{Map}(E,E') = \mathrm{Map}(\tilde O(X), \mathrm{Map}(E,E')) = \mathrm{Map}(X,\mathrm{Map}(E,E'))$$


Here $\tilde O(X)$ is the twisted arrow category of $X$, seen as an $\infty$-groupoid, which is just equivalent to $X$ (one way to see this is that the canonical projection $\tilde O(X)\to X$ is a left fibration with contractible fibers).


