One of the common methods of proving that an element $m$ is not definable in some structure $\frak M$, is to find an automorphism of $\frak M$ which moves $m$. However, this is not an equivalent condition, since it is possible to have a rigid structure with non-definable members (e.g. $\Bbb R$ as an ordered field, and "most" transcendental reals).
However, what happens when the structure is finite? Let me ask a specific question, but a more general answer is also welcomed.
Suppose that $\cal L$ is a language with a single binary relation symbol (and equality). If $M$ is a finite $\cal L$-structure with no automorphisms, is it true that every member of $M$ is definable? Is there some [nontrivial] sufficient and necessary condition to ensure pointwise definability of a finite $\cal L$-structure?
One variant which is also interesting to me, would be to slightly weaken and require only that some $m\in M$ is not moved by any automorphism, is it definable?
Let $\mathcal{L}$ be a finite language. If $M$ is a finite $\mathcal{L}$-structure, then an element $m\in M$ is definable if and only if it is fixed by all automorphisms of $M$. As a consequence, if $M$ has no nontrivial automorphisms, then every element of $M$ is definable.
Proof: Assume that $m$ is fixed by all automorphisms of $M$. Enumerate $M$ as $m = m_0,m_1,\dots,m_k$. Let $\varphi_M(m_0,\dots,m_k)$ be the $\mathcal{L}$-diagram of $M$. I claim that $\exists y_1,\dots,y_k\,\varphi_M(x,y_1\dots,y_k)$ defines $m$. Clearly $m$ satisfies this formula. And if $n\in M$ satisfies this formula, then letting $n_0 = n$ and $n_1,\dots,n_k$ the witnesses to the existential quantifiers, the map $m_i\mapsto n_i$ is an automorphism of $M$. Hence $m = n$. $\square$
By a similar argument, you can show that a subset of $M^n$ is definable if and only if it is fixed set-wise by all automorphisms of $M$. These facts are also true in arbitrary languages, using the usual sort of trick to reduce to a finite sublanguage. If you'd like me to say anything more along these lines, I can expand my answer.