Somewhat getting distracted from an exercise, I noticed that if a group $G$ has only one subgroup of order $p$ where $p$ is a prime (so our subgroup is wlog generated by $g$), any conjugation $$xgx^{-1}$$ must have order p as well, i.e. be representable as $g^k$ for $1\leq k < p$.
Question: Is there any simple criterion for the order of $xg$?
My observations
By induction, we can prove
Lemma Let $g, x\in G$ with $xg=g^kx$ with $1<k$. Then $x^ng = g^{k^n}x^n$, and furthermore,$$ (xg)^l = g^{k\left(k^l-1\right)/{(k-1)}} x^l $$
Now, if we demand $1<k<p$, we can note that \begin{align} e=g^{[\cdots]} &\iff p \vert k \frac{k^l-1}{k-1} \\ &\iff p \vert k \:\vee\: p\vert \frac{k^l-1}{k-1} \\ &\iff p(k-1)\vert k^l-1 \\ &\iff p \vert k^l-1 \:\wedge\: (k-1)\vert k^l-1 \\ &\iff p \vert k^l-1 \iff k^l \equiv 1 \mod p \\ &\iff l \vert p-1 \end{align} Where we used properties like $p$ being prime and $p$ and $k-1$ being coprime.
We get an immediate corollary of that:
Observation If $l\vert p-1$ and $\operatorname{ord}x\vert l$, we have $(xg)^l=e$, i.e. $\operatorname{ord} (xg) \mid l$.
Is there any way to strengthen this assertion? Assuming the best case scenario, can we get an explicit expression for the order of $xg$?
Proof of the commutation rule
First we'll prove $x^ng = g^{k^n}x^n$. The case $n=1$ is clear, so assume it holds for $n$. We have $$ x^{n+1}g = xx^ng = xg^{k^n}x^n = g^{k\cdot(k^n)}xx^n = g^{k^{n+1}}x^{n+1}. $$ Similarly, assume $(xg)^l = g^{k\left(k^l-1\right)/{(k-1)}} x^l$ (the base case $l=1$ is clear again). We have $$ (xg)^{l+1} = g^{k\left(k^l-1\right)/{(k-1)}} x^{l+1} g\\ = g^{k\left(k^l-1\right)/{(k-1)}} g^{k^{l+1}} x^{l+1}\\ = g^{k\left(k^l-1\right)/{(k-1)}} g^{k\left(k^{l+1}-k^l\right)/{(k-1)}} x^{l+1}\\ = g^{k\left(k^{l+1}-1\right)/{(k-1)}} x^{l+1}. $$