Not sure what you mean by "Jordan block matrix", but given that you are separating out scalar matrices, I expect you mean a matrix consisting of a single Jordan block associated to $\lambda$. In fact, it can be a matrix similar to a matrix with any number of Jordan blocks associated to $\lambda$, of any sizes that add up to $n$; the scalar matrix is a special case, in which you have $n$ $1\times 1$ Jordan blocks.
If $f^2=1_n$, then $f$ satisfies the polynomial $x^2-1$, hence the minimal polynomial of $f$ divides $x^2-1 = (x-1)(x+1)$. Since $\lambda$ is an eigenvalue if and only if $x-\lambda$ divides the minimal polynomial, the fact that there is a single eigenvalue says that the minimal polynomial is either $x-1$ or $x+1$, hence either $f=1_n$ or $f=-1_n$. However, since you say that the only eigenvalue is $\lambda=1$, that means that we must have $f=1_n$, and $f=-1_n$ is impossible.