Gfilui

Let $A$ be a symmetric square matrix with entries in $mathbbZ/pmathbbZ$ for a prime $p$ such that all of its diagonal entries are nonzero. Does there exists always a vector $x$ with all coordinates nonzero in the image of $A$? (this is true for $p=2$, but I don't know the answer for other values of $p$)

This is false. Let $c$ be a quadratic nonresidue modulo $p$. Our matrix will be $(p^2-1) times (p^2-1)$, with rows and colums indexed by pairs $(x,y) in mathbbF_p^2 setminus (0,0) $.

Our matrix is defined by
$$A_(x_1,y_1) (x_2, y_2) = x_1 x_2 - c y_1 y_2.$$
This is obviously symmetric. Since $c$ is a nonresidue, we have $x^2-cy^2 neq 0$ for $(x,y) in mathbbF_p^2 setminus (0,0) $, so the diagonal entries are nonzero.

Each column of this matrix is a linear function of $(x,y)$. So every vector in the image of this matrix is a linear function $mathbbF_p^2 setminus (0,0) longrightarrow mathbbF_p$ and, hence, takes the value $0$ somewhere.

We could make a smaller $(p+1) times (p+1)$ example by just taking one point $(x,y)$ on each line through $0$ in $mathbbF_p^2$.

Moreover, I claim that $(p+1) times (p+1)$ is optimal. In other words, if $A$ is an $n times n$ matrix with $n leq p$ and nonzero entries on the diagonal, then some vector in the image of $A$ has all coordinates nonzero. Interestingly, I don't need the symmetry hypothesis.

Let $W$ be the image of $A$. Note that $W$ is not contained in any of the coordinate hyperplanes.

Let $vecu= (u_1, u_2, ldots, u_n)$, among all elements of $W$, have the fewest $0$ entries. Suppose for the sake of contradiction that some $u_i$ is $0$. Then there is some other $vecv$ in $W$ with $v_i neq 0$. Consider the points $(u_j : v_j)$ in $mathbbP^1(mathbbF_p)$ as $j$ ranges over all indices where $(u_j, v_j) neq (0,0)$. There are fewer then $p+1$ such $j$, so some point of $mathbbP^1(mathbbF_p)$ is not hit, call it $(a:b)$. Then $-b vecu + a vecv$ is in $W$ and has fewer nonzero entries than $vecu$, a contradiction.