Let $A\subseteq X$. Is $$\mathrm{Aut}(X,A) \; = \; \Big\{ \varphi:X\to X\mid \text{$\varphi$ is bijective and $\varphi(a)\in A,\ \forall\,a\in A$} \Big\}$$ a subgroup of $({\rm Aut}(X),\circ)$? If not, show a counterexample.
I think it is, in fact, a subgroup. We just need to show that it has inverses and that it's closed.
Closed:
Let $\sigma,\tau\in\mathrm{Aut}(X,A)$. Since the composition of bijective functions is bijective, then it follows that $\tau\circ\sigma$ is bijective. Now, $\forall\,a\in A$, $$(\tau\circ\sigma)(a) = \tau(\sigma(a)) = \tau(a)=a\in A.$$ Hence, $\tau\circ\sigma\in\mathrm{Aut}(X,A)$.
Inverses:
Let $\sigma\in\mathrm{Aut}(X,A)$. Since $\sigma$ is bijective, then it's inverse $\sigma^{-1}$ is bijective. Now, $\forall\,a\in A$, $$ \sigma^{-1}(a) = \sigma^{-1}(\sigma(a)) = (\sigma^{-1}\circ\sigma)(a) = id(a) = a\in A. $$ Therefore, $\sigma^{-1}\in\mathrm{Aut}(X,A)$ and thus, $\mathrm{Aut}(X,A)$ it's a subgroup of $\mathrm{Aut}(X)$.
But since the problem asks if there is a counterexample, I'm not sure about my proof.
Fix $A\subseteq X$.
I will use the one-step subgroup test.
Clearly $H:={\rm Aut}(X,A)\subseteq G:={\rm Aut}(X).$
Since ${\rm id}_X$ is in $H$, we have $H\neq\varnothing$.
Let $\varphi, \theta\in H$. Let $a\in A$. Then $\varphi\circ\theta^{-1}$ is a bijection on $X$ by the standard properties of bijections. Also $a=\theta(b)$ for some $b\in A$ since $\theta|_A:A\to A$ given by $\theta|_A(c)=\theta(c)$ for all $c\in A$ is a bijection, so that
$$\begin{align} (\varphi\circ\theta^{-1})(a)&=\varphi(\theta^{-1}(a))\\ &=\varphi(b)\\ &\in A. \end{align}$$
Hence $\varphi\circ\theta^{-1}\in H$.
Hence $H\le G$.