Rotation about a coordinate axis is a linear transformation, so it has a correponding matrix. For instance, there is a matrix $A$ such that $A(x,y,z)^T$ is the vector that you get when you rotate $(x,y,z)^T$ $45^\circ$ about the $z$-axis. To find the matrix of a linear transformation, just work out where the transformation sends the basis vectors $(1,0,0)^T,(0,1,0)^T$, and $(0,0,1)^T$, and make those images the columns of the matrix.
For instance, the matrix that performs a $45^\circ$ rotation about the $z$-axis is $$A=\begin{bmatrix}1/\sqrt2&-1/\sqrt2&0\\1/\sqrt2&1/\sqrt2&0\\0&0&1\end{bmatrix}\;,$$ because this rotation takes $(1,0,0)^T$ to $(1/\sqrt2,1/\sqrt2,0)^T$, $(0,1,0)^T$ to $(-1/\sqrt2,1/\sqrt2,0)^T$, and $(0,0,1)^T$ to $(0,0,1)^T$.
Similarly, the matrix that performs a $90^\circ$ rotation about the $x$-axis is
$$B=\begin{bmatrix}1&0&0\\0&0&-1\\0&1&0\end{bmatrix}\;,$$ because that rotation takes the basis vectors $(1,0,0)^T,(0,1,0)^T$, and $(0,0,1)^T$ to $(1,0,0)^T$, $(0,0,1)^T$, and $(0,-1,0)^T$, respectively.
Scaling by $s_x,s_y,s_z$ means transforming $(x,y,z)^T$ to $(s_xx,s_yy,s_zz)^T$; this operation is performed by the matrix $$\begin{bmatrix}s_x&0&0\\0&s_y&0\\0&0&s_z\end{bmatrix}\;;$$ if you don’t immediately see why, calculate the vector $$\begin{bmatrix}s_x&0&0\\0&s_y&0\\0&0&s_z\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\;.$$
Translation by by $(2,1,-1)^T$ is not a linear transformation; to translate $(x,y,z)^T$ by $(2,1,-1)^T$, just add to get $(x+2,y+1,z-1)^T$.
Thus, if you start with a vector $(x,y,z)^T$ and perform the operations described in (1), you get the vector
$$B\left(SA\begin{bmatrix}x\\y\\z\end{bmatrix}+\begin{bmatrix}2\\1\\-1\end{bmatrix}\right)\;,\tag{1}$$
where $S$ is the appropriate scaling matrix.
Now you can simply substitute the vectors $t_0,t_1,t_2$, and $t_3$ for $(x,y,z)^T$, calculate their images under this sequence of transformations, and see whether these images are $u_0,u_1,u_2$, and $u_3$, respectively.
At worst you can do the same thing with each of (2)-(5); this is tedious, but it works. If you’ve already learned something about how rotations, scaling, and translations interact with one another, you may be able to shorten the work considerably. For example, if you know that rotation commutes with scaling, you’ll see that the operations in (1) and (4) have exactly the same result; in terms of $(1)$ this is because $SA=AS$, since $S$ is a diagonal matrix.
Best Answer
Hint:
Let's say you have a linear transformation $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$.
For convenience, denote $\mathbf{x}_i = T(\mathbf{e}_i)$, where $\mathbf{e}_i$ is the $i$th standard basis vector. So, for example, $\mathbf{e}_1 = \langle 1, 0, 0 \rangle$ in $\mathbb{R}^3$. Likewise, $\mathbf{e}_2 = \langle 0, 1, 0 \rangle$, etc.
Then $T$ is encoded by the $n \times n$ matrix $[\mathbf{x}_1, \mathbf{x}_2, \cdots, \mathbf{x}_n]$.
As a simple example, consider a $90^\circ$ counterclockwise rotation about the origin in $\mathbb{R}^2$. Note that $\langle 1, 0 \rangle \mapsto \langle 0, 1 \rangle$, and further $\langle 0, 1 \rangle \mapsto \langle -1, 0 \rangle$. So our linear transformation is encoded by the matrix $\left[ \begin{array}{ccc} 0 & -1 \\ 1 & 0 \end{array} \right]$.
Now applying this to solve your problem will be a bit of work, but if I'm not mistaken, the standard basis vectors lie on the midpoints of edges of this tetrahedron, and edges will map to edges. It'll be your task to figure out precisely where they are mapping to.
Another approach:
Let $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$, and $\mathbf{v}_4$ denote the vertices of your tetrahedron. Let $T$ be one of the given linear transformations. Then $T(\mathbf{v}_1) = \mathbf{v}_i$ for some $i$, and so forth. In this manner you can arrive at a system of equations whose solution yields the entries of the matrix encoding $T$.