Let $\rho: \mathbf{\Delta} \to \mathbf{\Delta}$ be the functor defined as the identity on objects and by the formula $$ \rho(f)(i)=n-f(m-i) $$ for any map $f:[m] \to [n]$, with $0 \leq i \leq m$. Then how to prove $\rho$ is the only non-trivial automorphism of the simplicial category $\mathbf{\Delta}$ ?
I only know any automorphism must map identities to identities, and I tried to use the decomposition of morphism but failed.
How to get started with the problem (maybe you can finish it):
Let $F : \Delta \to \Delta$ be an isomorphism of categories.
The object $[0] \in \Delta$ is the (only) terminal object in $\Delta$. It follows $F([0])=[0]$.
For $n \in \mathbb{N}$ the object $[n] \in \Delta$ has exactly $n+1$ morphisms $[0] \to [n]$. It follows that there are exactly $n+1$ morphisms $[0]=F([0]) \to F([n])$, and hence $F([n])=[n]$.
Now consider the two morphisms $\delta_0^1,\delta_1^1 : [0] \to [1]$, which we abbreviate by $\delta_0,\delta_1$ for the moment. The isomorphism $F$ maps them to the same set of morphisms, but their order may change. If the order stays the same, so $F(\delta_0)=\delta_0$, $F(\delta_1)=\delta_1$, we will prove $F = \mathrm{id}$ below. But if $F(\delta_0)=\delta_1$ and $F(\delta_1)=\delta_0$, consider $G := F \circ \rho$, where the automorphism $\rho$ is defined as in your question. It satisfies $\rho(\delta_0)=\delta_1$ and $\rho(\delta_1)=\delta_0$. But then $G(\delta_0)=\delta_0$ and $G(\delta_1)=\delta_1$. If we have proven $G=\mathrm{id}$, then $F = \rho^{-1}=\rho$.
Therefore, we may assume $F(\delta_0^1)=\delta_0^1$ and $F(\delta_1^1)=\delta_1^1$, and the goal is to prove $F(f)=f$ for every morphism $f : [m] \to [n]$.
The face maps in $\Delta$ are exactly the indecomposable monomorphisms in $\Delta$. And the degeneracy maps are exactly the indecomposable epimorphisms in $\Delta$. It follows that $F$ preserves these types of maps. And we know that they generate all morphisms under composition, so it suffices to look at these.
So for every $n > 0$ we find a permutation $\sigma_n$ of $\{0,\dotsc,n\}$ such that $F(\delta^n_i : [n-1] \to [n]) = \delta^n_{\sigma_n(i)} : [n-1] \to [n]$. (Similarly for the degeneracy maps.)
But these images still satisfy the simplicial identities. So for example, $$\delta^{n+1}_i \circ \delta^n_j = \delta^{n+1}_{j+1} \circ \delta^n_i$$ implies $$\delta^{n+1}_{\large \sigma_{n+1}(i)} \circ\delta^n_{\large \sigma_n(j)} = \delta^{n+1}_{\large \sigma_{n+1}(j+1)} \circ \delta^n_{\large \sigma_n(i)}.\quad (\ast)$$ The task is to derive $\sigma_n(i)=i$ from this relation as well as our assumption $\sigma_1 = \mathrm{id}_{\{0,1\}}$. I assume that we can simply compute both sides of $(\ast)$ to prove this. I suggest to start with proving $\sigma_2= \mathrm{id}_{\{0,1,2\}}$, and then to use an induction.