There are several ways I think of expressing this 'duality'. But before describing this, maybe it would help to explain a sense in which 'existence' (at least one element, or totality of a relation) is dual to 'uniqueness' (at most one element, or well-definedness of a relation).
So let's consider the category whose objects are sets and whose morphisms are relations $R: A \to B$: let $R(a, b)$ denote the truth value of $(a, b) \in R$, and for relations $R: A \to B$ and $S: B \to C$, define the composite $SR: A \to B$ by the rule $SR(a, c) = \exists_{b: B} R(a, b) \wedge S(b, c)$. The identity relation $1_A: A \to A$ is the diagonal subset of $A \times A$, defined by $1_A(a, b) \Leftrightarrow a = b$.
Technically this "category" is a 2-category, where 2-cells are given by inclusions $R \subseteq S$ between relations of the same type $A \to B$; we will write 2-cells as $R \leq S$. Even more, we have a $\dagger$-operation which takes a relation $R: A \to B$ to its opposite $R^{op}: B \to A$, where $R(a, b) \Leftrightarrow R^{op}(b, a)$. Altogether the structure one obtains is what is known as in categorical literature as an allegory, or as a bicategory of relations: there are various axiomatic frameworks for describing categories of relations.
There are also various notions of 'duality' in such a situation. One is by reversing the direction of 1-cells (called 'op'), another is by reversing the direction of 2-cells (called 'co'), and a third is by reversing directions of both (called 'co-op').
In this 2-categorical context, we may categorically express the condition that a relation $R: A \to B$ is well-defined or functional (for all $a \in A$ there exists at most one $b \in B$ such that $R(a, b)$ is true) by the condition $R \circ R^{op} \leq 1_B$.
The co-op dual of this condition is $1_A \leq R^{op} \circ R$, which translates to saying that to each $a \in A$ there exists at least one $b \in B$ such that $R(a, b)$, or that $R$ is a total relation.
If we have both conditions $R \circ R^{op} \leq 1_B$ and $1_A \leq R^{op} \circ R$, then the relation is a function; in other words, a relation $R$ is a function iff considered as a 1-cell in the 2-category $\mathbf{Rel}$, it has a right adjoint (which is necessarily $R^{op}$; this is a good exercise).
Onto the topology: if $X$ is a topological space and $\beta X$ is the set of ultrafilters on the underlying set of $X$, then there is a convergence relation $\gamma: \beta X \to X$ where $\gamma(U, x)$ (an ultrafilter $U$ converges to $x$) means that the filter of neighborhoods of $x$ is contained in $U$. In fact the very notion of topological space can be expressed in terms of ultrafilter convergence, as explored in the notion of 'relational $\beta$-module' for which you can find an account at the nLab.
A space $X$ is compact iff every ultrafilter on $X$ converges to at least one point. This is the same as saying that the convergence relation $\gamma: \beta X \to X$ is total. A space $X$ is Hausdorff iff every ultrafilter converges to at most one point. This is the same as saying that $\gamma: \beta X \to X$ is well-defined. A space is compact Hausdorff iff its convergence relation $\gamma$ is a function: this is an important ingredient in the theorem that compact Hausdorff spaces are exactly algebras of the ultrafilter monad.
Summarizing:
In the 2-category of sets and relations, the condition of compactness on the convergence relation of a topological space is co-op dual to the condition of Hausdorffness.
There are various other ways in which the duality between compactness and Hausdorffness manifests itself. One is that $X$ is compact if every projection map $X \times Y \to Y$ is closed, whereas $X$ is Hausdorff if the diagonal $X \to X \times X$ is closed (projections and diagonals being the two ingredients of product structures) -- although it would take some time to elaborate a sense in which these properties should be seen as "dual".
There is an example which satisfies something much stronger: there exist nontrivial groups $G$ such that any homomorphism (not even necessarily continuous) $\pi:G\to GL_n(k)$ for any field $k$ (and, in fact, any commutative integral domain) is trivial, so in particular for any $a,b\in G$ we have $\pi(a)=\pi(b)$. Since any group can be given a locally compact Hausdorff topology, namely the discrete topology, these will in particular answer your question.
As for examples of such groups $G$, we can take any finitely generated which has no finite quotients, e.g. the Higman group mentioned in the comment by Terry Tao.
This result is apparently due to Mal'cev, but I don't have a reference at hand so here is a sketch of an argument. It is enough to show that if you have a finitely generated subgroup $\Gamma$ of a group $GL_n(R)$ for some integral domain $R$ (e.g. the image of $G$ under a representation), then $\Gamma$ is residually finite (hence trivial, if $\Gamma$ is a quotient of $G$ which has no finite quotients). The main idea of the proof is to replace $R$ by some ring which is finitely generated over $\mathbb Z$, meaning it is a quotient of a polynomial algebra over $\mathbb Z$. Then $R$ is a Jacobson ring, so in particular the intersection of all maximal ideals of $R$ is trivial, and moreover for all maximal ideals $m$ of $R$ we have $R/m$ finite. Now if $\gamma\in\Gamma$ is any nontrivial element, then there is a maximal ideal $m$ not containing all coefficients of $\gamma$, so $\gamma$ has nontrivial image in the finite group $GL_n(R/m)$.
Best Answer
Yes, it is "standard" that a finite-dimensional vector space over a complete, non-discrete, division algebra (!) has a unique topology compatible with a topological vector space structure (=Hausdorff, addition is continuous, scalar multiplication is continuous). This is probably proven (at least for real or complex scalars, but the proofs should generalize) in any source on general TVS's, e.g., section 5, my notes. Without completeness, or without discreteness, there are easy counter-examples.