Translation functor in a triangulated category under certain hypotheses Let $\mathcal{T}$ be a $\Bbbk$-linear triangulated category which is Hom-finite and Krull-Schmidt, with translation functor $\Sigma$ satisfying $\Sigma^2 = \text{id}$. Suppose that $\mathcal{T}$ has finitely many indecomposables
$$ U_1^{(i)}, U_2^{(i)}, \dots , U_{n_i}^{(i)}$$
for $i = 1, \dots , m$, where $m,n_1, \dots n_m \in \mathbb{N}$. Write $\mathcal{F}_i = \{ U_1^{(i)}, \dots , U_{n_i}^{(i)} \}$. Suppose further that 
$$ \text{Hom}_{\mathcal{T}} ( U_k^{(i)} , U_\ell^{(j)}) = 0$$
whenever $i \neq j$, while
$$ \text{Hom}_{\mathcal{T}} ( U_k^{(i)} , U_\ell^{(i)}) \neq 0$$
for all $i,k, \ell$.

Does it follow that restricting $\Sigma$ to the elements of $\mathcal{F}_i$ induces a permutation of $\mathcal{F}_i$? That is, for each $i,k$, does there exist $\ell$ with $\Sigma U_k^{(i)} = U_\ell^{(i)}$?

It's easy to see that $\Sigma$ induces a permutation of the $U_k^{(i)}$ and that, for each $i$, $\Sigma \mathcal{F}_i \subseteq \mathcal{F}_j$ for some $j$, with equality only if $\mathcal{F}_i$ and $\mathcal{F}_j$ have the same size. Therefore, if the $\mathcal{F}_i$ have sizes which are pairwise distinct then the result holds.
However, I run into trouble when $|\mathcal{F}_i| = |\mathcal{F}_j|$ for some $i \neq j$. I can't even do the case when there are four indecomposables which split into two families $\mathcal{F}_1$ and $\mathcal{F}_2$. My current approach has been to consider long exact sequences arising from applying certain $\text{Hom}$ functors to triangles, but this has only got me so far.
 A: There's a trivial counterexample with two indecomposables $U$ and $V$, where $U$ and $V$ have endomorphism ring $\mathbb{k}$, $\operatorname{Hom}(U,V)=0$, $\operatorname{Hom}(V,U)=0$, $\Sigma U=V$, $\Sigma V=U$, and the only distinguished triangles are finite direct sums of
$$U\stackrel{\text{id}}{\to}U\to 0\to \Sigma U=V$$
and rotated versions.
But this is essentially the only counterexample. Suppose $\mathcal{F}_i$ is a family of indecomposables as in the question, with $\Sigma\mathcal{F}_i\subseteq\mathcal{F}_j\neq\mathcal{F}_i$, let $\alpha:X\to Y$ be a nonzero map between two objects of $\mathcal{F}_i$ and complete to a triangle
$$X\stackrel{\alpha}{\to}Y\stackrel{\beta}{\to}Z\stackrel{\gamma}{\to}\Sigma X.$$
Then the restriction of $\gamma$ to every indecomposable summand of $Z$ must be non-zero, or else the triangle has a nontrivial direct sum decomposition, contradicting the indecomposability of $X$ and $Y$ and the fact that $\alpha\neq0$. So every indecomposable summand of $Z$ is isomorphic to an object of $\mathcal{F}_j$, since it has a non-zero map to $\Sigma X$, and therefore $\beta=0$. But then $\alpha$ is a split epimorphism, and since $X$ is indecomposable, $\alpha$ must be an isomorphism.
Therefore $\mathcal{F}_i$ has only one object $X$, up to isomorphism, with $\operatorname{End}(X)$ a division ring.
