$\newcommand\v\vec\newcommand\R{\mathbb R}$For any natural $k\ge2$, $k$ inequalities (but not fewer than $k$) will suffice.
Indeed, take any vector $\v a\in(0,\infty)^k$ such that $\v1\cdot\v a=1$, where $\v1:=(1,\dots,1)\in\R^k$ and $\cdot$ denotes the dot product. Take any linearly independent vectors $\v g_1,\dots,\v g_{k-1}$ in $\R^k$ that are orthogonal to $\v a$, and then let $\v g_k:=-\v g_1-\dots-\v g_{k-1}$. Then, for each $\v x\in\R^k$, the inequalities
\begin{equation}
\v g_1\cdot\v x\ge0,\ \dots,\ \v g_{k-1}\cdot\v x\ge0,\ \v g_k\cdot\v x\ge0 \tag{0}
\end{equation}
imply the equalities
\begin{equation}
\v g_1\cdot\v x=\dots=\v g_{k-1}\cdot\v x=0. \tag{1}
\end{equation}
Also, the vector $\v1$ is linearly independent of $\v g_1,\dots,\v g_{k-1}$ -- otherwise, we would have $1=\v1\cdot\v a=0$, since (1) holds for $\v x=\v a$. So, the system of equalities (1) together with the equality
\begin{equation}
\v1\cdot\v x=1 \tag{2}
\end{equation}
has a unique solution.
Moreover, equalities (1) and (2) do hold for $\v x=\v a$.
Thus, $\v x=\v a$ is the unique solution of the system of inequalities (0) and equality (2),
as desired.
Also, for any natural $k\ge2$, any $k-1$ inequalities of the form
\begin{equation}
\v g_1\cdot\v x\ge0,\dots,\v g_{k-1}\cdot\v x\ge0 \tag{0a}
\end{equation}
will not suffice to identify $\v a$, even together with the equality (2).
Indeed, suppose the contrary, so that $\v a$ is the only solution in $(0,\infty)^k$ of the system (0a)--(2).
Consider first the case when $\v a$ is not orthogonal to
\begin{equation*}
V:=\text{span}(\v g_1,\dots,\v g_{k-1}).
\end{equation*}
Take any nonzero vector $\v h\in\R^k$ orthogonal to $V$ and such that $\v1\cdot\v h\ge0$, so that $\v1\cdot(\v a+\v h)\ge1>0$, and also short enough so that $\v a+\v h\in(0,\infty)^k$. Let
\begin{equation}
\v b:=\frac{\v a+\v h}{\v1\cdot(\v a+\v h)}\in(0,\infty)^k. \tag{3}
\end{equation}
Then $\v b\ne \v a$, while (0a) and (2) hold with $\v x=\v b$. This contradicts $\v a$ being the only solution in $(0,\infty)^k$ of the system (0a)--(2).
Next, consider the case when $\v a$ is orthogonal to $V$ and $\v g_1,\dots,\v g_{k-1}$ are linearly dependent, so that there is a nonzero vector $\v h\in\R^k$ orthogonal to $V$ and to $\v a$ and such that $\v1\cdot\v h\ge0$, which also is short enough so that $\v a+\v h\in(0,\infty)^k$. Then again, for $\v b$ as in (3), we have $\v b\ne \v a$, while (0a) and (2) hold with $\v x=\v b$. This contradicts $\v a$ being the only solution in $(0,\infty)^k$ of the system (0a)--(2). Again, a contradiction.
Finally, consider the case when $\v a$ is orthogonal to $V$ and $\v g_1,\dots,\v g_{k-1}$ are linearly independent. Then there is a vector $\v h_1\in V$ such that $\v g_1\cdot\v h_1=1$ and $\v g_j\cdot\v h_1=0$ for $j=2,\dots,k-1$. Letting now $\v h:=t\,\v h_1$ for a small enough $t>0$ so that $\v1\cdot\v h\ge-1/2$ and $\v a+\v h\in(0,\infty)^k$, we get $\v1\cdot(\v a+\v h)\ge1/2>0$.
So, again for $\v b$ as in (3), we have $\v b\ne \v a$, while (0a) and (2) hold with $\v x=\v b$. This contradicts $\v a$ being the only solution in $(0,\infty)^k$ of the system (0a)--(2). Once again, a contradiction.
Thus, $k$ is the smallest number of inequalities of the form $\v g_j\cdot\v x\ge0$ that will suffice.
