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.
Autoequivalences preserve categorical properties. Therefore, they preserve homotopy limits and colimits, hence the zero object, cofibers, sums, and so on.
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 $Y \to Y \to \dots \to Y \to 0 \to 0 \to \dots$.
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.
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.
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 $end(X) \otimes R \simeq end_R(R \otimes X)$.
Let's define the indicator $g(n,m)$ by $$ g(n,m)_k \simeq \begin{cases} \Bbb S &\text{if }n \leq k \neq m,\\0 &\text{otherwise,}\end{cases} $$ with all transition maps being identities. Note that $end(g(n,m)) \simeq \Bbb S$. (I'll allow $m=\infty$ in this definition.)
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 $Y$ is has one such summand: it is a shift of an indicator.
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 5, and hence it is an indicator by point 6. 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 $g(n,m) \to X$.
As a result, the family $\{\Sigma^r g(n,m)\}$ is preserved (up to equivalence) by any autoequivalence.
Now we note that $$ map(\Sigma^r g(n,m), \Sigma^s g(p,q)) \simeq \begin{cases} S^{s-r} &\text{if }p \leq n \leq q < m,\\ S^{s-r-1} &\text{if }n < p \leq m \leq q,\\ 0 &\text{otherwise.} \end{cases} $$
Therefore, suppose $g$ is equivalent to an indicator and satisfies the following property: for any other indicator $h$, $map(g,h) \simeq \Bbb S$ if and only if $g \simeq h$. Then $g$ is equivalent to $\Sigma^r g(0,\infty)$ for some $r$. Therefore, any autoequivalence must take $g(0,\infty)$ to something equivalent to $\Sigma^r g(0,\infty)$ for some $r$.
Similarly, for any other inducator $h$, $map(h, \Sigma^r g(0,\infty)) \simeq \Bbb S$ if and only if $h \simeq \Sigma^r g(n,\infty)$ for some $n$.
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 $map(h_1,h_2) \simeq \Bbb S$.
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 $$ \Sigma^r g(0,\infty) \leftarrow \Sigma^r g(1,\infty) \leftarrow \Sigma^r g(2,\infty) \leftarrow \Sigma^r g(3,\infty) \leftarrow \dots $$
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.