Does a degeneration always have a larger-dimensional automorphism group? 
Suppose $\newcommand{\X}{\mathcal{X}}\X$ is an algebraic stack over a field $k$, $\xi$ is a $k$-point which has another $k$-point $x$ in its closure ($x$ is an isotrivial degeneration of $\xi$). Must the dimension of $Aut(x)$ be larger than the dimension of $Aut(\xi)$?

If $\X=[X/G]$, where $G$ is an algebraic group over $k$, then the answer is yes. An isotrivial degeneration corresponds to a $G$-orbit with another $G$-orbit in its closure. The closure of a $G$-orbit can only contain smaller dimensional orbits, and the dimension of the orbit is complementary to the dimension of the stabilizer of a point in that orbit, so when one orbit degenerates to another, there is always a jump in the dimension of the stabilizer.
Almost every algebraic stack I can think of is étale-locally of the form $[X/G]$, and étale maps preserve the dimensions of automorphism groups. This suggests that the answer is probably "yes."

Example
Consider the 1-parameter family of $2\times 2$ matrices $\begin{pmatrix}1&t\\ 0&1\end{pmatrix}$. If we are studying matrices up to conjugation, then away from $t=0$, the family is (isomorphic to) the constant family $\begin{pmatrix}1&1\\ 0&1\end{pmatrix}$. However, at $t=0$ you get a different Jordan type. So we say that this family is an isotrivial degeneration of Jordan types $\begin{pmatrix}1&1\\ 0&1\end{pmatrix}\rightsquigarrow \begin{pmatrix}1&0\\ 0&1\end{pmatrix}$. When you pass to the more degenerate Jordan type, the automorphism group (i.e. the group of ways in which the matrix is self-conjugate) jumps from something 2-dimensional to something 4-dimensional.

Motivation
It sometimes happens that you want to determine all degeneration relations among a collection of points in an algebraic stack. For example, you may be trying to determine if some map is weakly proper.† If the answer to this question is "yes," then you can rule out certain degenerations by looking at dimensions of automorphism groups.
†See Weakly proper moduli stacks of curves by Alper, Smyth, and van der Wyck.
 A: This is not a complete answer but a suggestion towards a "yes" answer, too long for a comment. 
You can represent $\xi$ and $x$ by a morphism $\varphi: S\to\mathcal{X}$, where $S$ is a normal affine $k$-scheme of finite type with a point $s\in S(k)$, such that $\varphi(s)=x$ and $\varphi_{\mid U}$ factors through $\xi$, where $U:=S\setminus\{s\}$. On $S\times S$, consider the two objects $\varphi_1$ and $\varphi_2$ of $\mathcal{X}$ obtained by pullback via the projections. Put $Y:=\underline{\mathrm{Isom}}_{\mathcal{X}}(\varphi_1,\varphi_2)$. This is an algebraic space of finite type over $S\times S$, and a pseudo-torsor under the group we are interested in. The assumptions on $x$ and $\xi$ imply that the image of $f:Y\to S\times S$ is $(U\times U)\cup\{(s,s)\}$. Hence $f$ is not open, and therefore not equidimensional by EGA IV (14.4.4) since $S\times S$ is geometrically unibranch. Hopefully, this implies the result on the group but we have to be careful; for instance our group may have constant dimension but extra components at $x$, which would account for the lack of equidimensionality at some points of the fibre at $(s,s)$ (but not all?).
