Note: This is not a full answer. This is intended to introduce a mental framework to help anyone parse the existing varied answers to this questions.
(0.A) Before you start reading: Are you on an educational program, studying for a course, or preparing for an examination? If this is the case, the institution that is in charge of your study is also in charge of deciding what answers would be accepted. Therefore, your best bet is to stick to what they have decided, rather than looking for answers elsewhere. This is true even if their answers are wrong. You simply parrot the wrong answer that they would accept, while reminding yourself not to stick to this answer once you finish your study.
(0.B) The type of computer systems also matter. For most people (students and software programmers), the two types that matters are: "general purpose computers" and "embedded systems".
If you need answers for "embedded systems", users of the Electronics Stack Exchange will offer more relevant answers. Be warned that today's "embedded systems" are moving towards biscuit-sized PCs, which means a lot of practitioners nowadays are using miniature general-purpose computers to do the work of what embedded systems used to do. In embedded systems of the last decade, hobbyist circuit boards may contain discrete electronic components, such as "32k x 8-bit SRAM", etc. These are the truly "byte-addressable" memory running on 4-bit or 8-bit bus. The current trend of using miniature general purpose computers imply that those discrete electronic components may become a thing of the past.
If you need answers for, say, "mainframe computers", or computer architectures that have since been discontinued, please refer to Retrocomputing Stack Exchange.
If you need answers for "8-bit computing", both sites (Electronics; Retrocomputing) will offer useful advice.
(1) The answer as seen from a C programmer: see gnasher729's answer. Note that, the C standard's definition for a byte is that it can hold "at least 256 different values", implying at least 8 bits (or equivalents), without requiring that it is exactly 8 bit.
(2) The answer as seen from a programmer who writes assembly language or machine code (the binary code that the CPU would execute): They will see what the Instruction Set Architecture (ISA) has specified for. The x86 ISA allows for instructions that updates one byte at a time. However, there is no guarantee that updating a single byte (8 bits) would take one-eighth the time for updating a 64-bit value. (It is very well possible that updating an 8-bit value in memory could take the same amount of time as updating a 64-bit value in memory.)
(3) The answer as seen on the CPU cache: see user253751's answer, first paragraph. It may require understanding of cache coherence protocols. Refer to the Wikipedia article for more information. On Intel architectural documents, the performance of CPU memory bus is measured in "transactions" ("MT/s": mega-transactions per seconds, "GT/s" giga-transactions per seconds). A "transaction" refers to the minimum unit of data transfer per command.
(4) The answer as seen on the memory bus: SDRAM bus commands are issued by the memory controller to the RAM modules. You will need to read the section burst ordering and burst length in order to understand a technically accurate answer. On SDRAM (and later standards, such as DDR4), "word size" is 64-bit. See Wikipedia article on this topic: Synchronous_dynamic_random-access_memory@Commands
(5) The answer as seen on the electrical control circuits that reside on the SDRAM module: see Martin Maat's answer, second paragraph.
(6) The answer as seen on rows of transistor-based memory cells on the memory chip: see user253751's answer, second paragraph. An entire row of memory cells are updated at once. ("Row" is the commonly used terminology when referring to SRAM and DRAM. "Block" is the commonly used terminology when referring to non-volatile random-access-memory, NVRAM. They both refer to the smallest organization of memory units that must be updated all at once.)
(7) The answer as seen from the Operating System (OS). As long as an application is accessing its own memory, and it is not the first time the application touches the memory page, the OS will not contravene. Normally, the answers (1 - 6) will apply as usual. However, several situations will require OS intervention: (1) the application has asked the OS to give it a fresh new empty page, the OS has granted it (and gave it the memory page address to that new page), but has yet to prepare it. In this case, the first time memory access will trigger some page preparation operations. This may involve writing more than one byte of data to the memory. (2) the application has tried to access some memory address it is not allowed to. In this case, the OS will stop it.