Consider a quadratic program
$$\min_{x} x^TQx + b^tx$$
such that $Ax\leq c$ pointwise. This is a quadratic program with linear inequality constraints. Under what conditions for the data (matrix $Q$, vector $b$, vector $c$ and matrix $A$) do we have uniqueness of the solution? Let's assume that $Q$ is symmetric positive semidefinite and not equal to the $0$ matrix. Similarly we can assume $A$ is not the $0$ matrix. I will also assume that the constraint set is compact, but I am unwilling to assume that $Q$ is positive definite.
Idea: A sufficient condition will be that, at a solution $x^\star$, we want the set of feasible directions and the kernel of $Q$ to intersect trivially. This obviously cannot happen if there is a solution in the interior of the constraint set unless $Q$ is simply positive definite. So, I would also be interested in conditions that ensure that the solution is on the boundary of the constraint set.
Edit: The correct notion for this is to consider the tangent cone to the constraint set $\{Ax\leq c\}$. If the tangent cone intersects with the kernel of $Q$ in a trivial way (i.e., $=\{0\}$) then the solution must be unique.
I have an answer that you may consider partial, I can't give explicit conditions of the entries but at least some conditions that, I think, you can verify automatically with a computer. Also, I assume that you found $x^*$ and you want a condition to verify its uniqueness.
Reduction to some case where calculs might be easier
First of all, you can always bring back to the case where $Q = \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}$ and $b = \delta_{r + 1}$ (i.e. $b_k = 1$ if $k = r + 1$, $0$ else) or $0$.
Indeed, let us begin by showing that we can choose $b \in \ker(Q)$. As $Q$ is symmetric, $\ker(Q) \oplus \mathrm{im}(Q)$ is the whole $\mathrm{R}^n$ and this decomposition is orthogonal. Let $b = Qb_1 + b_2$ with $b_2 \in \ker(Q)$. Notice that, $$ f(x) := x^\top Qx + bx = \left(x + \frac{1}{2}b_1\right)^\top Q\left(x + \frac{1}{2}b_1\right) + b_2^\top x - \frac{1}{4}b_1^\top Qb_1. $$ The term $- \frac{1}{4}b_1^\top Qb_1$ is just a constant and doesn't influence where the minimum of $f$ is and replacing $x$ by $y = x + \frac{1}{2}b_1$, we can also replace $b$ by $b_2 \in \ker(Q)$. The new condition on $y$ is that $Ay = Ax + \frac{1}{2}Ab_1 \leqslant c + \frac{1}{2}Ab_1$. Replace $c$ by $c + \frac{1}{2}Ab_1$.
Assume from now that $b \in \ker(Q)$. Let us show that we can choose $Q = \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}$ and $b = \varepsilon\delta_{r + 1}$ where $\varepsilon \in \{0,1\}$. Indeed, by the spectral theorem, we can write $Q = \Omega^\top D\Omega$ with $D$ of the form $\mathrm{diag}(\lambda_1,\ldots,\lambda_r,0,\ldots,0)$ where $r = \mathrm{rk}(Q) < n$. $b \in \ker(Q)$, which means that $\Omega b$ is of the form $(0,\ldots,0,*,\ldots,*)$ with the $r$ first coefficients equal zero. $\mathrm{GL}_{n - r}(\mathbb{R})$ acts transitively on $\mathbb{R}\backslash\{0\}$ so we can find an $(n - r) \times (n - r)$ invertible matrix $\tilde{R}$ such that $\begin{pmatrix} 0 & 0 \\ 0 & \tilde{R} \end{pmatrix}\Omega b = \varepsilon\delta_{r + 1}$ with $\varepsilon \in \{0,1\}$ in function of if $b = 0$ or not.
Let $R = \begin{pmatrix} \mathrm{diag}\left(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_r}\right) & 0 \\ 0 & \tilde{R}^\top \end{pmatrix}$ and $x = \Omega^\top Ry$. We have, $$ f(x) = y^\top R^\top\Omega Q\Omega^\top Ry + b^\top\Omega^\top Ry = y^\top R^\top DRy + (R^\top\Omega b)^\top y = y^\top\begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}y + \varepsilon\delta_{r + 1}^\top y. $$ It proves the wanted result. The new condition on $y$ is $Ax = A\Omega^\top R y \leqslant c$ so replace $A$ by $A\Omega^\top R$.
To be honest, I don't even know if this reduction is useful but it can help to simplify some computations and in this case, you have $$ f(x) = \sum_{k = 1}^r y_k^2 + \varepsilon y_{r + 1}. $$ Notice that $\varepsilon = 0$ if and only if $b \in \mathrm{im}(Q)$ in the original problem.
the cone $C$
Now, consider $K$ the compact polyhedron defined by $Ax \leqslant c$. If $x^* \in K$ is a solution to your optimisation problem, let $C$ be the set of all $v$ such that $x^* + \varepsilon v \in K$ for $\varepsilon > 0$ small enough. Notice that $C$ is a closed cone of $\mathbb{R}^n$ and $C = \mathrm{R}^n$ if and only if $x \in \overset{\circ}{K}$. By convexity of $K$, $x^* + \varepsilon v \in K$ implies that for all $\varepsilon' < \varepsilon$, $x^* + \varepsilon'v \in K$. If we call $a_k^\top$ the rows of $A$, the conditions $Ax \leqslant c$ means that for all $k$, $a_k^\top x \leqslant c_k$. In this case, let $I$ be the set of all $i$ such that $a_i^\top x^* = c_i$. In this case, the equation of $C$ is $\forall i \in I, a_i^\top v \leqslant 0$, hence the equation of $C + x^*$ is $\tilde{A}x \leqslant \tilde{c}$ where $\tilde{c} = (c_i)_{i \in I}$ and $\tilde{A} = \left(a_i^\top\right)_{i \in I}$.
Let us show that $x^*$ is a minimum of $f$ on $C + x^*$. Indeed, for all $v \in C$, there is a $\varepsilon > 0$ such that $x^* + \varepsilon v \in K$ hence, $$ 0 \leqslant f(x^* + \varepsilon v) - f(x^*) = 2\varepsilon x^{*\top}Qv + \varepsilon^2v^\top Qv + \varepsilon b^\top v, $$ and without loss of generality, $\varepsilon < 1$ so, $$ f(x^* + v) - f(x^*) = 2x^{*\top}Qv + v^\top Qv + b^\top v = \frac{1}{\varepsilon}(f(x^* + \varepsilon v) - f(x^*)) + (1 - \varepsilon)v^\top Qv, $$ and this quantity is non-negative. $x^*$ is indeed a minimum of $f$ on $C$. Notice that if $f(x^* + v) = f(x^*)$, then $f(x^* + \varepsilon v) - f(x^*) = 0$ for all $0 < \varepsilon < 1$ small enough and $v^\top Qv = 0$. The second equality is equivalent to $v \in \ker(Q)$ because $Q$ is positive semidefinite and the first one implies then that $b^\top v = 0$ (this will be useful later). All this implies that for all $t \geqslant 0$, $f(x^* + tv) = f(x^*)$.
Main result
$x^*$ is the unique minimum of $f$ in $K$ if and only if it is unique in the whole $C$ if and only if $C \cap \ker(Q) \cap \mathrm{Span}(b)^\top = 0$. The first equivalence is a consequence of the fact that for all $v \in C$, $f(x^* + v) = f(x^*) \Rightarrow \forall t \geqslant 0, f(x^* + tv) = f(x^*)$. Moreover, we saw that if $f(x^* + v) = f(x^*)$, then $v \in \ker(Q) \cap \mathrm{Span}(b)^\top$. It proves the converse sens of the second equivalence.
For the direct sens, assume that there exists a $v \in C \cap \ker(Q) \cap \mathrm{Span}(b)^\top$ that is not zero. Then $$ f(x^* + v) - f(x^*) = 2x^{*\top}Qv + v^\top Qv + b^\top v = 0. $$ It proves the equivalence.
Remark : $\ker(Q) \cap \mathrm{Span}(b)^\top$ always have dimension $n - r - 1$ unless $b \in \mathrm{im}(Q)$ (i.e. $b = 0$ if you choose $b \in \ker(Q)$), and $C \cap \ker(Q) \cap \mathrm{Span}(b)^\top = 0$ is equivalent to $K \cap \ker(Q) \cap \mathrm{Span}(b)^\top = 0$ but you get rid of some superfluous equations when you go from $K$ to $C$.
I hope it helps !