Are n-truncated quasicategories a model for n-categories?

Let $C$ be an $\infty$-category, and $n\geq -1$. The following are equivalent:

1. $C$ is $n$-truncated.

2. The $\infty$-groupoids $\def\Map{\operatorname{Map}}\Map(\Delta^0,C)$ and $\Map(\Delta^1,C)$ are $n$-truncated. (Remember that $\Map(B,C)$ is the maximal Kan complex inside $\operatorname{Fun}(B,C)$.)

3. ($n\geq0$ only) For all pairs of objects $x,y$ in $C$, the $\infty$-groupoid $\Map_C(x,y)$ is $n$-truncated and $\def\Aut{\operatorname{Aut}}\Aut_C(x)$ is $(n-1)$-truncated. (Here $\Aut_C(x)\subseteq \Map_C(x,x)$ is the subobject of self-equivalences.)

(1) $\Rightarrow$ (2) is immediate. (2) $\Rightarrow$ (1) is because $\mathrm{Cat}_\infty$ is generated under colimits by $\Delta^0$ and $\Delta^1$.

(2) $\Leftrightarrow$ (3) is via the fiber sequences $$ \Map_C(x,y) \to \Map(\Delta^1,C) \to \Map(\Delta^0,C)\times \Map(\Delta^0,C). $$ associated to each pair of objects $(x,y)$ and $$ \Aut_C(x) \to * \to \Map(\Delta^0,C) $$ associated to each object $x$ (This last one is why you need $n\geq 0$.)

I think it is wrong. Consider an $\infty$-category $C$ with two objects $x,y$ such that $Map(x,y)=X$ for some $n$-truncated space (which is not ($n-1$)-truncated) and all other mapping spaces are trivial (i.e. $Map(y,x)$ is empty and the other two contain only the identity) and the compositions are the obvious ones.

Now, I claim that for $n\ge 0$ the category $C$ is $n$-truncated as an object of $Cat_{\infty}$, but it is not an $n$-category. The second claim is clear since $X$ is not ($n-1$)-truncated. To check the first claim, we need to show that $Map(K,C)$ is $n$-truncated for every $\infty$-category $K$ (where $Map$ is the maximal $\infty$-subgroupoid of the $\infty$-category of functors). It is enough to check this for $K=\Delta^n$, since every $K$ is a colimit of simplices, $Map(-,C)$ turns colimits into limits and $n$-truncated spaces are closed under limits. Moreover, it is enough to check this only for $K=\Delta^0, \Delta^1$ using a similar argument and the "Segal conditions". Finally, for $K=\Delta^0$ we get a discrete 4 element set and for $K=\Delta^1$ we get the disjoint union of the 4 mapping spaces which is also $n$-truncated.