Does finite(ish) language imply upwards-directed potential automorphism groups? This is a follow-up to this question. Briefly, say that two structures $\mathfrak{A},\mathfrak{B}$ are parametrically equivalent, and write "$\mathfrak{A}\approx\mathfrak{B}$," iff $(i)$ they have the same underlying set and $(ii)$ each primitive function/relation in one is definable in the other (allowing parameters) - note that $(ii)$ is equivalent to saying that each has the same set of definable-with-parameters relations of each arity.
Given a structure $\mathfrak{A}$, I'm interested in the "automorphism group spectrum" $\mathsf{AGS}(\mathfrak{A})=\{Aut(\mathfrak{B}): \mathfrak{B}\approx\mathfrak{A}\}$. At the above-linked question, Harry West showed that $\mathsf{AGS}(\mathfrak{A})$ need not be upwards-directed: even if $\mathfrak{A}\approx\mathfrak{B}$ there need be no $\mathfrak{C}\approx\mathfrak{A}$ such that $Aut(\mathfrak{A})\cup Aut(\mathfrak{B})\subseteq Aut(\mathfrak{C})$. However, that argument crucially used an infinite language.
My question is whether we can still get failures of upwards-directedness:

Is there a structure $\mathfrak{A}$ in a finite language (or even a language with only finitely many non-constant symbols - call this "almost finite language") such that $\mathsf{AGS}(\mathfrak{A})$ is not upwards-directed?

Note that if $\mathfrak{X}\approx\mathfrak{Y}$ and $\mathfrak{Y}$ has almost finite language, then there is some $\mathfrak{X}'\approx\mathfrak{X}$ with almost finite language such that $Aut(\mathfrak{X}')=Aut(\mathfrak{X})$. So arguably restricting to the almost finite language case is a natural thing to do, in a way that looking at actually finite languages isn't (the trivial group is always in $\mathsf{AGS}(\mathfrak{A})$ but this requires adding infinitely many constant symbols in general).
 A: No to the title question, yes to the question in the post.$\DeclareMathOperator{Aut}{Aut}$ Pick an infinite set $X.$ Pick infinite coinfinite sets $Y_A,Y_B\subseteq X$ with $Y_B=Y_A\cup\{y\}$ for some $y\not\in Y_A.$ Define $\mathfrak A$ and $\mathfrak B$ to have

*

*underlying set $X$

*a unary predicate $U,$ interpreted as follows

*$U^\mathfrak A(z)$ iff $z\in Y_A$

*$U^\mathfrak B(z)$ iff $z\in Y_B$
These have the same definable sets.
Suppose for contradiction that there is a structure $\mathfrak C$ with $\mathfrak C\approx\mathfrak A$ and
$\Aut(\mathfrak A)\cup\Aut(\mathfrak B)\subseteq \Aut(\mathfrak C).$
$\Aut(\mathfrak A)$ is the group of symmetries fixing $Y_A$ setwise, and $\Aut(\mathfrak B)$ is the group of symmetries fixing $Y_B$ setwise.
$\Aut(\mathfrak C)$ contains every transposition $\pi$ swapping an element $x\in Y_A$ with an element $x'\in X\setminus Y_A.$ I will try to actually describe the symmetry this time. Recall $y\in Y_B\setminus Y_A.$ Conjugate $\pi$ by a transposition in $\Aut(\mathfrak A)$ to reduce to the case $x'=y.$ Then use a transposition in $\Aut(\mathfrak B)$ to swap $x$ and $x'.$  (More generally, $\Aut(\mathfrak C)$ contains all permutations $\pi$ of $X$ such that $|\pi(Y_A)\setminus Y_A|$ and $|Y_A\setminus \pi(Y_A)|$ are finite and of equal cardinality.)
By the assumption $\mathfrak C\approx\mathfrak A$ there is a sequence of parameters $y_1,\dots,y_k\in X$ and a relation symbol $R$ such that
$$z\in Y_A\iff R^{\mathfrak C}(z,y_1,\dots,y_k).$$
Let $Y_0=\{y_1,\dots, y_k\}.$
Pick $x\in Y_A\setminus Y_0$ and $x'\in X\setminus (Y_A\cup Y_0).$ We must have $R^{\mathfrak C}(x,y_1,\dots,y_k)$ and $\neg R^{\mathfrak C}(x',y_1,\dots,y_k).$ But there is a symmetry of $\Aut(\mathfrak C)$ swapping $x$ and $x’$ and fixing all elements of $Y_0.$ This contradicts the choice of $\mathfrak C.$
