Homogeneous Suslin tree in $L$ (Exercise 15.27 of Jech's Set Theory) Exercise 15.27 of Jech's Set Theory says:

If $V = L$ then there exists a homogeneous Suslin tree.

Recall that a Suslin tree $T$ is a tree of height $\omega_1$ such that every chain and antichain is at most countable. $T$ is furthermore homogeneous if for every $x,y \in T$ on the same level, there exists an automorphism $\pi : T \to T$ (i.e. $s,t$ are compatible iff $\pi(s),\pi(t)$ are compatible).
It is implied by the previous exercises that we need to use the fact that $V = L$ implies $\diamondsuit$. Indeed, Theorem 15.26 shows that the existence of a $\diamondsuit$-sequence, $\langle S_\alpha : \alpha < \omega_1\rangle$ implies that a Suslin tree exists, by asserting that if $S_\alpha$ is an antichain of $T_\alpha$ ($T$ restricted to branches of length $<\alpha$) where $\alpha$ is limit, then let $T_{\alpha+1}$ extend $T_\alpha$ such that $S_\alpha$ remains an antichain in $T_{\alpha+1}$. The exact construction of $T_{\alpha+1}$ is highlighted in Lemma 15.25.
I suspect that the approach is as follows: We inductively shows that $T_\alpha$ is homogeneous for all $\alpha$. The only non-trivial step is $T_{\alpha+1}$, where $\alpha$ is limit. However, I fail to see how the construction of $T_{\alpha+1}$ highlighted in Lemma 15.25 allows $T_{\alpha+1}$ to remain homogeneous. I'm also not sure for the case where $S_\alpha$ is not an antichain, so the construction of $T_{\alpha+1}$ is not given by Lemma 15.25, how one can construct $T_{\alpha+1}$ to make it homogeneous.
 A: Okay here is a rough attempt but it's too long for a comment and does not use $\Diamond'$. Fix $b$ a cofinal branch and let $\mathcal{A}$ be a group of autormorphisms that act transitively on the levels of $T_\alpha$ and is countable. We have that for each $x\in T_\alpha$ there is a $\pi\in \mathcal{A}$ such that $x\in \pi(b)$ I define then for each $\pi\in \mathcal{A}$ a unique point $t_\pi\in T_\alpha$ that is the limit of the branch $\pi(b)$. Since  a group acts transitively on itself we have that $T_{\alpha+1}$ is homogeneous. The way it was built means that $S_\alpha$ is bounded as an antichain since every $a$ there is a $\pi(b)$ that passes through it and $t_\pi$ bounds it. If $S_\alpha$ is not an anti chain the construction is still valid.
Edit adding somethings that may not be obvious. The existence of $\mathcal{A}$ can be shown by picking for each pair of $x,y$ that lie on the same level an automorphism. This is countable and so the subgroup generated will also be countable.
Also Jech requires normal trees to have a bunch of conditions the only conditions you need for the tree to have so that lemmas 15.24,15.25 still hold is that for each $x$ the set $\{y>x\}$ is not linearly ordered and intersects all levels of the tree above $x$.
