Given are two intersecting lines $l$ and $l'$ in $\mathbb{E}^2$. How does one show that there are exactly four isometries that map $l$ to $l'$ and have $l\cap l'$ as fixed point?
Intuitively, I've managed to come up with three; rotations about $l \cap l'$ (clockwise and counterclockwise), and reflection about their bisection. I haven't got a clue as to how to find a fourth one, let alone formally proving these results.
Imagine any circle around the point of intersection. It will intersect $l$ in $A_1$ and $A_2$ and $l'$ in $B_1$ and $B_2$.
If you swap $l$ and $l'$ by an isometry, you have to exchange $\{A_1,A_2\}$ and $\{B_1,B_2\}$. Which in general (see below) means either $A_1\leftrightarrow B_1,A_2\leftrightarrow B_2$ or $A_1\leftrightarrow B_2,A_2\leftrightarrow B_1$. These four pairs between preimage and image point uniquely define the transformation, as long as $l$ and $l'$ are distinct. Geometrically speaking, the transformations are the reflections in the two angular bisectors between the lines.
You mention rotations, but a rotation that maps $l$ to $l'$ won't map $l'$ to $l$ unless the two lines happen to be orthogonal (or identical). So I'd exclude these. I think for the situation you described there can only be two isometries with the required properties, so what you want to show is in fact wrong. If you want to consider the orthogonal case, then you'd get defining points like $A_1\mapsto B_1\mapsto A_2\mapsto B_2\mapsto A_1$ and $A_1\mapsto B_2\mapsto A_2\mapsto B_1\mapsto A_1$. But in the non-orthogonal case, $\lVert A_1-B_1\rVert\neq\lVert B_1-A_2\rVert$ so the transformation would not be an isometry.
So what's wrong with the question? I can only guess. Perhaps you are to show that there can be no more than 4 isometries, so you indeed have to consider the orthogonal case. Or you don't have to swap $l$ and $l'$, but instead should fix $\{l,l'\}$ so that you also have to consider operations which map $l$ to $l$ and $l'$ to $l'$. In that case, you'd have the identity transformation and a rotation by $180°$ as additional isometries. Have a look at your problem statement, and see whether one of these alternative interpretations could be what's actually being asked.