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

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.

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

Therefore, suppose $g$ is equivalent to an indicator and satisfies the following property: (EDIT: had to correct this property) for any other indicator $h$, if $map(g,h)$ and $map(h,g)$ are both nontrivial, then $h \simeq \Sigma^k g$ for some $k$. 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 $\Sigma^r$.

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

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.

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 $g(n,m) \to X$.

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

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

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.

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 $n$.

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 $\Sigma^r$.

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

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.

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: (EDIT: had to correct this property) for any other indicator $h$, if $map(g,h)$ and $map(h,g)$ are both nontrivial, then $h \simeq \Sigma^k g$ for some $k$. 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 $\Sigma^r$.

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

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.

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

Therefore, suppose $g$ is equivalent to an indicator and satisfies the following property: (EDIT: had to correct this property) for any other indicator $h$, if $map(g,h)$ and $map(h,g)$ are both nontrivial, then $h \simeq \Sigma^k g$ for some $k$. 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 $\Sigma^r$.

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

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 $\Sigma^r$.

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

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.

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: (EDIT: had to correct this property) for any other indicator $h$, if $map(g,h)$ and $map(h,g)$ are both nontrivial, then $h \simeq \Sigma^k g$ for some $k$. 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 $\Sigma^r$.