The first question was already answered David Roberts and Jonathan Chiche. Let me address the second one. It's not reasonable to expect that such a model structure exists. We can ask instead whether there is a model structure on simplicial sets in which nerves of fibrant categories (in Thomason's model structure) are fibrant. And, in fact, such a model structure does exist. We can just transfer the Quillen model structure on simplicial sets along the double subdivision functor. Note that the Thomason's model structure can be now transferred from this new model structure. Also, this model structure on simplicial sets presents the homotopy theory of spaces since weak equivalences in it are just ordinary weak equivalences.
Now, just for completeness, let me answer the question in the post. Of course, there are are a lot of model structures on simplicial sets in which the nerves of categories are fibrant. For example, we can take the left Bousfield localization of the Joyal model structure with respect to the maps $\Delta^n \amalg_{\Lambda^n_k} \Delta^n \to \Delta^n$. The trivial model structure also satsfies the conditions. These model structures do not present the homotopy theory of spaces. If we want to keep the class of weak equivalences the same as in the Quillen model structures, then I can show that there is no model structure in which nerves of categories are fibrant. Actually, I will prove a stronger statement:
If there is a model structure on simplicial sets in which $\Delta^1$ is fibrant and contractible, then its homotopy category is thin.
First, note that if $f : X \to Y$ is a weak equivalence between fibrant objects and $A$ is cofibrant, then $f$ has the weak right lifting property with respect to $A$. That is, for every map $g : A \to Y$, there is a map $g' : A \to X$ such that $f \circ g'$ is homotopic to $g$.
Now, let $A$ be a cofibrant simplicial set in the hypothetical model structure. Consider the inclusion of the left endpoint $f : \Delta^0 \to \Delta^1$. It is a weak equivalence between fibrant objects by the assumptions. Let $g : A \to \Delta^1$ be the constant map at the other endpoint. Then the observation in the previous paragraph implies that $g$ is homotopic to $g'$, the constant map at the left endpoint. Let $A \amalg A \to C(A)$ be a cylinder object for $A$. The previous observation implies that there are no 1-simplices in $C(A)$ between two components (if there is, then $g$ and $g'$ cannot be homotopic).
Thus, any cylinder object $A \amalg A \to C(A)$ equals to $s_1 \amalg s_2 : A \amalg A \to A_1 \amalg A_2$ for some maps $s_1 : A \to A_1$ and $s_2 : A \to A_2$. Moreover, there are retractions $r_1 : A_1 \to A$ and $r_2 : A_2 \to A$ of $s_1$ and $s_2$, respectively. Thus, two maps $f,g : A \to B$ are homotopic if and only if $f$ factors through $s_1$ and $g$ factors through $s_2$. But, since $s_1$ and $s_2$ have retractions, all maps factor through them, so any two maps are homotopic. This implies that any two maps in the homotopy category are equal.