$ f(x) = \delta_{ x: 1^Tx = k} $ is an indicator function that return 0 if x is in $\{ x: 1^Tx = k\} $ and infinity otherwise.
I'm calculating its convex conjugate $ f^*(z) = \sup_x z^Tx - f(x) = \delta_0(z) $ because $f^*(z)$ goes to infinity when $z \ne 0$.
I know that $f^{**}=f $ because f is closed and convex, so I computed the convex conjugate of $f^*(z)$ which is
$f^{**}=f =\sup_z z^tx - \delta_0(z) $
but now it seems that this expression is just zero because z must be zero otherwise it goes to -infinity, and I could not recover $f$
Can someone tell me what am I doing wrong?
You computation of $f^*$ is wrong. Call $\pi:\Bbb R^n\to\mathbf 1^\perp$ the orthogonal projection and $q:\Bbb R^n\to\Bbb R$ the functional such that $q(v)\mathbf 1=v-\pi(v)$. I.e. $q(v)=\frac1n\mathbf 1^\top v$. $$f^*(z)=\sup\{z^\top x\,:\, \mathbf 1^\top x=k\}=\sup\{q(z)k+\pi(z)^\top x\,:\, \mathbf 1^\top x=k\}$$
Now, if $\pi(z)\ne 0$, then indeed $f^*(z)=\infty$ because we can evaluate it on the sequence $x_m=m\pi(z)+\frac kn\mathbf 1$. But if $\pi(z)=0$, then $f^*(z)=q(z)k=\frac kn\mathbf 1^\top z$.
So actually $f^*(z)=\frac kn \mathbf 1^\top z+\delta_{\operatorname{span}\mathbf 1}(z)$.