TITLE: Conjugate subgroup strictly contained in the initial subgroup?
QUESTION [65 upvotes]: Let $G$ be a group, $H\subseteq G$ a subgroup and $a\in G$ an element of the group. Is it possible that $aHa^{-1} \subset H$, but $aHa^{-1} \neq H$? 
If $H$ has finite index or finite order, this is not possible.

REPLY [7 votes]: This is a repost of my answer to another incarnation of this question (An example of a group, a subgroup and an element, satisfying a given condition.). I will delete the answer from there and just leave it here since it seems better to have all answers gathered in one place.
Here is a general way to construct such groups and elements:
Let $X$ be a set with $X = Y\cup Z \cup W$, such that $Y$, $Z$ and $W$ are pairwise disjoint, and such that $|X| = |Y| = |Y\cup Z| = |W|$. 
Fix a bijection $\varphi: Y\cup Z\to Y$ and a bijection $\psi: W\to Z\cup W$.
Define a new bijection $\chi: X\to X$ by $\chi(x) = \varphi(x)$ for $x\in Y\cup Z$ and $\chi(x) = \psi(x)$ for $x\in W$.
Let $H$ be the subgroup of the group of bijections from $X$ to $X$ consisting of those elements that fix everything in $Y$.
Now it is easy to check that $\chi^{-1}H\chi$ is the subgroup consisting of those elements that fix everything in $Y\cup Z$, which is a strictly smaller subgroup than $H$.
The very nice example given by Mikko Korhonen is a special case of this, with $X = \mathbb{Z}$, $Y$ the set of non-positive integers, $Z = \{1\}$ and $W$ the set of integers strictly greater than $1$.