You didn't make the precise assumptions on $f$ explicit, but even in a fairly general situation, your $\widehat{f}_1$ is the only example that is supported by $[-1,1]$.

Given an $f$ with the stated properties, we can let $\widehat{g}=\widehat{f}^{1/2}$, so $f=g*g$, and if $\textrm{supp}\, f\subseteq [-1,1]$, then Titchmarsh's convolution theorem gives that $g$ is supported by $[-1/2,1/2]$. Since we're dealing with real valued functions $f,\widehat{f}$ here, they are even, and thus $$ 1=f(0) = \int_{-1/2}^{1/2} g(x)g(-x)\, dx = \int_{-1/2}^{1/2} g(x)^2\, dx \ge \left(\int_{-1/2}^{1/2} g(x)\, dx\right)^2 = \widehat{g}(0)^2=1 . $$ Equality in the Cauchy-Schwarz inequality means that the functions $1,g$ are linearly dependent, so $g=1$, and this gives us the triangular function $f(x)=1-|x|$ as the only possible example supported by $[-1,1]$.

You didn't make the precise assumptions on $f$ explicit, but even in a fairly general situation, your $\widehat{f}_1$ is the only example that is supported by $[-1,1]$.

Given an $f$ with the stated properties, we can let $\widehat{g}=\widehat{f}^{1/2}$, so $f=g*g$, and if $\textrm{supp}\, f\subseteq [-1,1]$, then Titchmarsh's convolution theorem shows that $g$ is supported by $[-1/2,1/2]$. Since we're dealing with real valued functions $f,\widehat{f}$ here, they are even, and thus $$ 1=f(0) = \int_{-1/2}^{1/2} g(x)g(-x)\, dx = \int_{-1/2}^{1/2} g(x)^2\, dx \ge \left(\int_{-1/2}^{1/2} g(x)\, dx\right)^2 = \widehat{g}(0)^2=1 . $$ Equality in the Cauchy-Schwarz inequality means that the functions $1,g$ are linearly dependent, so $g=1$ (or $=-1$), and this gives us the triangular function $f(x)=1-|x|$ as the only possible example supported by $[-1,1]$.