Not an answer, but too long for a comment.

About your Example: If I understand right, you want to consider the self-equivalence of the category of simplicial spectra given by the automorphism of $\Delta$ that reverses the order in a finite ordered set. The corresponding self-equivalence of the category of nonnegative graded chain complexes just takes a chain complex and switches the sign of the boundary map in even degrees. That's naturally isomorphic to the identity. (At least, for ordinary chain complexes, and I suppose that that is basically true for spectral chain complexes as well.)

And here's a side comment: what if we look at spectral chain complexes of some definite dimension? For example, consider the category of 1-dimensional complexes $X_1\to X_0$ (that is, complexes with $X_n=0$ for all $n>1$). This is really just the category of maps of spectra. There is a self-equivalence of the category of these:

$F(X_1\to X_0)=(X_0\to X_0/X_1)$

We have $F^2(X_1\to X_0)=(X_0/X_1\to \Sigma X_1)$ and

$F^3(X_1\to X_0)=(\Sigma X_1\to \Sigma X_0)$.

So the group of (homotopy classes of) self-equivalences has in it not just the infinite cyclic group generated by suspension, but this larger group generated by a cube root of suspension. (Maybe that's all there is?)

What I just noticed is that something similar happens for higher-dimensional complexes. For two dimensions we are looking at objects $(X_2\to X_1\to X_0)$, where this really means three spectra, two maps, and a nullhomotopy of the composed map $X_2\to X_0$; or in other words a map $X_2\to X_1$ and a map $X_1/X_2\to X_0$.

Write $T(X)$ for the cofiber of that map $X_1/X_2\to X_0$; it is also the cofiber of the corresponding map $\Sigma X_2\to X_0/X_1$. I am thinking of it as the homology of the spectral chain complex. Here is a functor, an auto-equivalence of this category of $2$-dimensional spectral chain complexes:

$F(X_2\to X_1\to X_0)=(X_1\to X_0\to T(X))$

$F^2(X_2\to X_1\to X_0)=(X_0\to T(X)\to \Sigma^2 X_2)$

$F^3(X_2\to X_1\to X_0)=(T(X)\to \Sigma^2 X_2\to \Sigma^2 X_1)$

$F^4(X_2\to X_1\to X_0)=(\Sigma^2 X_2\to \Sigma^2 X_1\to \Sigma^2 X_0)$

So this time we have a fourth root of $\Sigma^2$.

It looks like in general for $d$-dimensional nonnegatively graded spectral chain complexes we have a self-equivalence that is a $(d+2)$nd root of $\Sigma^d$.

I don't know quite what to make of this, or whether it is useful at all in answering your question.