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If a key $d$ is used to encrypt a message $m$ that is identical to the key $d$, resulting in a cipher $c$

$c = \text{Encrypt}(m, d) = \text{Encrypt}(d, d)$

is the computational cost to brute force $d$ lower than if the key $d \ne m$ ?

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Depends on the encryption scheme.

In the IND-CPA (or even IND-CCA) security game, the adversary is not given the secret key $d$ (obviously) and hence cannot ask for $\mathrm{Enc}(d,d)$.

In other words: an encryption scheme where $\mathrm{Enc}(d,d) = d$ for all keys $d$ can still be IND-CPA(/CCA) secure (but they'd be completely useless for your scenario).

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    $\begingroup$ This is related to the concept of Circular Security... $\endgroup$ – Hilder Vítor Lima Pereira Feb 19 '18 at 15:57

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