Recall that if $Y$ is a contracticle space, and $X$ is path-connected, then the set of classes of homotopic maps $Y\rightarrow X$ is trivial. That is, every map is null-homotopic and all null-homotopic maps are homotopic to each other (because $X$ is path-connected). Given that $[0,1]$ is contractible, and your cylinder $X$ is path-connected, what does this tell you about $f,g,h,j$?
The question asker has mentioned that he is using homotopy relative to a subspace, in which case the above doesn't hold. Note that the Cylinder $S^1\times [0,1]$ is homotopy equivalent to the circle $S^1$ via the standard projection map. This projection map also has the nice property that each of your paths become loops. That is, $p\circ f(0)=p\circ f(1)$ and similarly for $g,h$ and $f*j$ where $p$ is the homotopy equivalence given by projection.
Now let $X$ and $Y$ be homotopy equivalent spaces and $f_1\colon X\rightarrow Y$ and $f_2\colon Y\rightarrow X$ be such that $f_1f_2\simeq 1_Y$ and $f_2f_1\simeq 1_X$. A path $\gamma\colon [0,1]\rightarrow X$ is homotopic to a path $\lambda\colon [0,1]\rightarrow X$ relative to end points if and only if $f_1\circ\gamma$ is homotopic to $f_1\circ\lambda$ relative to end points.
In your case, $Y$ is a circle, $f_1$ is projection on to the base circle and so your paths are only homotopic relative to end points if the induced loops on the circle are homotopic loops relative to base point. Hopefully you know how to test if loops in the circle are homotopic relative to base point (look at the winding number) and so you should be able to tell that only $g$ and $f*j$ are homotopic paths relative to base point.