RSA exhibits symmetrical properties — is it possible to form a cryptanalytic attack?
The first observation in the question boils down to: in textbook RSA, the encryption of $N-M$ instead of $M$ yields $N-C$ instead of $C$.

This is a special case of a more general property of textbook RSA, that the encryption of $M\cdot M'\bmod N$ yield $C\cdot C'\bmod N$, whenever enciphering $M$ yields $C$ and enciphering $M'$ yields $C'$; combined with the elementary fact that enciphering $M'=N-1$ yields $C'=N-1$.

These properties are the basis of numerous attacks on textbook RSA, including some of these (and of attacks on real RSA signature schemes with ad-hoc padding, such as on ISO/IEC 9796-2 scheme 1). This is why textbook RSA should not be directly used for anything else than random messages, or messages sufficiently close to that.



A second observation in the question considers the case of $N=P\cdot Q$ with $P-1$ smooth.

The worst thing if that happens is that it is easy to factor such $N$, using Pollard's p-1 factoring algorithm (giving a total break, making the considerations in the question of secondary importance). However this is not a serious weakness of RSA, for the odds of accidentally selecting such a $P$ are vanishingly small for usual order of magnitude of $P$; see the Dickman function for a numerical estimate.

Further, we can avoid such $P$ with relative ease; this remains standard practice at least for $P$ up to 512 bits (see appendix B.3 in FIPS 186-4, and in particular the requirement in B.3.1 A: "p and q with lengths of 512 bits shall be generated using the conditions based on auxiliary primes"). The only case when this might matter (there is not quite a consensus here) is when one generates ziiillions of keys, with $\log_2P$ below or close to 512, and a marginal risk that even one public key is vulnerable to Pollard's p-1 factoring is unacceptable (the adversary's best use of CPU time might be to attack each key with Pollard's p-1, rather than a few keys with other factoring methods).