Are Convolution and Deconvolution Kernels the Same?
Question - When Are the Convolution Operator (Kernel) and the Deconvolution Operator (Kernel) the Same?

Discrete Convolution (Cyclic) is described by Circulant Matrix.
Circulant Matrices are diagonalized by the DFT Matrix (Will be denoted as $ F $).
So given a convolution kernel $ g $ with its matrix form given by $ G $ one could have its diagonalization by:

$$ G = {F}^{H} \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) F $$

Where $ \mathcal{F} \left( g \right) $ is the Discrete Fourier Transform (DFT) of the kernel $ g $.

In the general case the deconvolution is given by:

$$ x = G h \Rightarrow {G}^{-1} x = {G}^{-1} G h = h $$

Where $ {G}^{-1} $ is the Deconvolution Operator (Inverse of $ G $ the convolution operator).
In our case we're looking for the case $ G = {G}^{-1} \Rightarrow G G = I $.  

Utilizing the diagonalization of Discrete (Cyclic) Convolution Operators by the DFT Matrix (Which is Unitary Matrix):

$$ G G = {F}^{H} \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) F {F}^{H} \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) F = {F}^{H} \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) F $$

Namely we need $ \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) = I $ which means $ \forall i, \, {\mathcal{F} \left( g \right)}_{i} \in \left\{ -1, 1 \right\} $, namely the Fourier Transform of $ g $ has the values of -1 or 1.
So there is a set of kernels $ \mathcal{S} $ which is defined by:

$$ \mathcal{S} = \left\{ g \mid \forall i, \, {\mathcal{F} \left( g \right)}_{i} \in \left\{ -1, 1 \right\} \right\} $$

One should notice that the identity Kernel ($ I $) is indeed part of this set of kernels.

The above kernels obey $ \delta \left[ n \right] = \left( g \circ g \right) \left[ n \right] $ where $ \circ $ is the Circular Convolution (Which is multiplication of the DFT).

How could one generate such a kernel?
Well, just generate a vector of numbers which each is composed of $ 1 $ or $ -1 $ and use its ifft().
Pay attention that if you are after real operator the vector must obey the symmetry of the DFT for Real Numbers.

You may find a code to replicate the above in my StackExchange Signal Processing Q19646 GitHub Repository.