Here is a proof that any autoequivalence of (the $\infty$-category of) filtered spectra is naturally equivalent to a suspension functor. Since filtered spectra are equivalent to simplicial spectra, this translates back. The argument takes some work but the following sketches it.

1. Autoequivalences preserve categorical properties. Therefore, they preserve homotopy limits and colimits, hence the zero object, cofibers, sums, and so on.

2. In particular, autoequivalences preserve _compact_ objects. A filtered spectrum $X_0 \to X_1 \to \dots$ is compact if and only if each $X_i$ is compact and the maps $X_n \to X_{n+1}$ are equivalences for $n >> 0$. This condition is sufficient because such objects are built in finitely many stages from free objects $0 \to \dots \to 0 \to S^n \to S^n \to \dots$, and you can check necessity by mapping out to diagrams of objects of the form $Y \to Y \to \dots \to Y \to 0 \to 0 \to \dots$.

3. Because filtered spectra are stable, they get mapping spectra as part of the categorical structure. If $X$ and $Y$ are compact then $map(X,Y)$ is a finite spectrum.

4. Similarly because filtered spectra are a cocomplete, stable $\infty$-category, they also get a left-tensoring over spectra: $Z \otimes (X_n) \simeq (Z \otimes X_n)$. Any autoequivalence must preserve this left-tensoring.

5. For any filtered spectrum $X$, we get a ring spectrum $end(X)$, and if $X$ is compact then by the above we find that for any ring spectrum $R$ we can calculate $end(X) \otimes R \simeq end_R(R \otimes X)$.

6. Let's define the _indicator_ $g(n,m)$ by
$$
g(n,m)_k \simeq \begin{cases} \Bbb S &\text{if }n \leq k \leq m,\\0 &\text{otherwise,}\end{cases}
$$
with all transition maps being identities. Note that $end(g(n,m)) \simeq \Bbb S$. (I'll allow $m=\infty$ in this definition.)

7. For any field $k$, any filtered $Hk$-module $Y$ is a direct sum of shifts of indicators: $\Sigma^r Hk \otimes g(n,m)$. As a result, $end_{Hk}(Y) \simeq Hk$ if and only if if $Y$ has one such summand: it is a shift of an indicator.

8. Here is a lemma: a compact filtered spectrum $X$ satisfies $end(X) \simeq \Bbb S$ if and _only_ if it is equivalent to a suspension $\Sigma^r g(n,m)$ for some $r, n, m$. To check this, we note that for any field $k$, the filtered $Hk$-module $Hk \otimes X$ satisfies $end_{Hk}(Hk \otimes X) \simeq Hk$ by point 6, and hence it is an indicator by point 7. For this to be true, the universal coefficient theorem shows that $H\Bbb Z \otimes X$ must a shifted indicator $\Sigma^r H\Bbb Z \otimes g(n,m)$, and the Hurewicz / Whitehead theorems then let us construct an equivalence $g(n,m) \to X$.

9. As a result, the family $\{\Sigma^r g(n,m)\}$ of (shifted) indicators is preserved (up to equivalence) by any autoequivalence.

10. Now we note that
$$
map(\Sigma^r g(n,m), \Sigma^s g(p,q)) \simeq \begin{cases}
S^{s-r} &\text{if }p \leq n \leq q \leq m,\\
S^{s-r-1} &\text{if }n < p \leq m + 1, $m < q,\\
0 &\text{otherwise.}
\end{cases}
$$

11. I claim that property 10 allows us to identify $\Sigma^r g(0,\infty)$ among indicators, and therefore any autoequivalence must take $g(0,\infty)$ to something equivalent to $\Sigma^r g(0,\infty)$ for some $r$. In previous edits I've taken shortcuts with mixed success; here is a lengthier argument.
* Let's define an _essential_ map to be a map $f: g \to h$ between indicators such that $map(g,h) \simeq \Bbb S$, generated by $f$. By the above, essential maps are all equivalent to (shifts of) standard maps $g(n,m) \to g(p,q)$ for $p \leq n \leq q \leq m$ and maps $g(n,m) \to \Sigma (p,q)$ for $n < p \leq m+1$, $m < q$.
* We will say that an essential map is _bounded_ if there is an upper bound on the number of inequivalent ways to write it as a composite of essential maps. The maps $g(n,m) \to g(p,q)$ are unbounded if $q < m = \infty$, and otherwise they are bounded; the maps $g(n,m) \to \Sigma g(p,q)$ are all unbounded unless $q=\infty$.
* Therefore, shifts of $g(n,\infty)$ are the only objects with no bounded maps out and only bounded maps in.
* If there is an essential map $g(n,\infty) \to g(m,\infty)$, then $m \leq n$, and so shifts of $g(0,\infty)$ are uniquely determined among these objects.

12. Similarly, for any other indicator $h$, $map(h, \Sigma^r g(0,\infty)) \simeq \Bbb S$ if and only if $h \simeq \Sigma^r g(n,\infty)$ for some $n$.

13. If $h_1 \simeq \Sigma^r g(n_1,\infty)$ and $h_2 \simeq \Sigma^r g(n_2,\infty)$ are two such indicators as in point 11, we have that $n_1 \geq n_2$ if and only if $map(h_1,h_2) \simeq \Bbb S$.

14. This lets us show that, up to equivalence, any autoequivalence of this category must, up to equivalence, take the tower
$$
g(0,\infty) \leftarrow g(1,\infty) \leftarrow g(2,\infty) \leftarrow g(3,\infty) \leftarrow \dots
$$
to a tower equivalent to
$$
\Sigma^r g(0,\infty) \leftarrow \Sigma^r g(1,\infty) \leftarrow \Sigma^r g(2,\infty) \leftarrow \Sigma^r g(3,\infty) \leftarrow \dots
$$

15. This shows that, up to equivalence, the autoequivalence is equivalent to the shift $\Sigma^r$ on the subcategory spanned by $g(n,\infty)$. However, this is a set of generators for the category of filtered spectra: every other filtered spectrum is built from them by hocolims. Therefore, the autoequivalence must be equivalent to the shift $\Sigma^r$.

As a final note, there is an equivalence of categories between $Fun^L(Fil(Sp), Fil(Sp))$ and the functor category $Fun(\Bbb N \times \Bbb N^{op}, Sp)$ by a compact-generators argument. Therefore, the _space_ of autoequivalences can be determined by looking at the space of self-maps of the object representing the identity functor in this category.