Let $K$ be the global function field $\mathbb{F}_3(t)$ and set $$ \Gamma_1(t^2)=\left\{\begin{pmatrix} a & b\\ c & d \end{pmatrix}\in \text{GL}_2(\mathbb{F}_3[t]) : a\equiv 1 \pmod{t^2} \text{ and } c\equiv 0\pmod{t^2}\right\}.$$ This group acts on $\mathbb{P}^1(K)=K\cup \{\infty\}$ by fractional linear transformation: For $r/s\in \mathbb{P}^1(K)$, $$\begin{pmatrix} a & b\\ c & d \end{pmatrix} \frac{r}{s}=\frac{ar+bs}{cr+ds}.$$
The cusps of the Drineld modular curve $X_1(t^2)$ correspond with the orbits of the above action, $\Gamma_1(t^2)\backslash \mathbb{P}^1(K)$. I computed that $$\{\infty=1/0,\ 0=0/1,\ 1/t,\ 1/(t+1),\ 1/(t+2)\}$$ is a set of representatives for the cusps. However, according to Proposition 6.6(i) of this paper (Invariants of Some Algebraic Curves Related to Drinfeld Modular Curves by Gekeler), there should only be 4 cusps. In particular, two of the cusps in my set should be equivalent, but I have not been able to see how this is possible. Can anyone explicitly show me that two of my cusps are in the same orbit of $\Gamma_1(t^2)$?
(I also computed the cusps of $X_0(t^2)$, finding them to be $\{\infty,\ 0,\ 1/t\}$. In this case, the number of cusps I found agrees with the corresponding formula in the paper of Gekeler mentioned above. This suggests that $1/(t+1)$ and $1/(t+2)$ are likely the equivalent cusps in $X_1(t^2)$.)
I think that the formula is wrong and that there are at least five cusps in $X_1(t^2)$, those that you highlighted.
Indeed, it’s not too hard to see that $\Gamma_1(t^2) \cdot \infty$ contains only rational fractions with at least a double pole at $t=0$ (so none of the other representatives); $\Gamma_1(t^2) \cdot 1/t$ contains fractions with a pole at $t$; $\Gamma_1(t^2) \cdot 0$ contains rational fractions defined at $0$. But if $\begin{bmatrix}a & b\\c&d\end{bmatrix}0=\frac{1}{t\pm 1}$, as $b,d$ are coprime, $d=\pm (t \pm 1)$ and thus (determinant!) $a = \pm (t\pm 1)$ mod $t^2$, a contradiction.
So the only possible equality of cusps is for $1/(t+1)$ and $1/(t+2)$. But let’s prove it doesn’t happen.
Assume that there is some $M \in \Gamma_1(t^2)$ such that $M\frac{1}{t+1}=\frac{1}{t+2}$. Then $N=\begin{bmatrix} 1& 0\\ -(t+2)&1\end{bmatrix}M\begin{bmatrix}1&0\\t+1&1\end{bmatrix}$ maps $\infty$ to itself so is of the form $\pm \begin{bmatrix}\epsilon & u \\0&1\end{bmatrix}$, where $u \in \mathbb{F}_3[t]$ and $\epsilon=\pm 1$.
Then $\begin{bmatrix} 1& 0\\ (t+2)&1\end{bmatrix} \begin{bmatrix}\epsilon & u \\0&1\end{bmatrix}\begin{bmatrix}1&0\\-(t+1)&1\end{bmatrix}=\pm M \in \pm \Gamma_1(t^2)$.
But then, one can compute that $\pm M=\begin{bmatrix}1&0\\t+2&1\end{bmatrix}\begin{bmatrix}v & u\\-t-1&1\end{bmatrix}=\begin{bmatrix}v & w \\ (t+2)v-(t+1) & x\end{bmatrix}$ where $v=\epsilon -u(t+1)$, and $w,x \in \mathbb{F}_3[t]$. We thus need that $v \pm 1$ be divisible by $t^2$ and that $(t+2)v-(t+1)$ too be divisible by $t^2$, and we easily check this to be impossible.