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I have a couple users who need multiple email addresses (alias) forwarded to their accounts but setting up a separate user and forwarding the email is showing the email to the users main address instead of the aliased email.

For example, here is what I need: john.doe@work.com is the users email address but they also need to receive emails sent to redundancy@work.com, and redundant.department@work.com. When the emails are sent to the other two email addresses I want them to be forwarded to the user and showing that it was sent to the redundancy email address and not john.doe because the user needs to be able to filter these emails as well as reply from the redundancy department email address.

How can I set up alias' in exchange 2010 to work this way?


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3 Answers 3

Being able to reply from one of your aliases in Exchange is a question that gets asked a lot, and unfortunately there is no easy solution, despite people asking for it since Exchange 2000. As far as MS is concerned email addresses you have setup as aliases are just a way to route email to you, and that's it. Take a look at this question that talks about being able to send from alternative addresses.

In your situation, what I would suggest is to set up separate mail accounts for these addresses you want to send and receive from, and give the user who needs to access them full permissions, including send as rights. They can then add these mailboxes as additional mailboxes in Outlook, so they can operate them through a single Outlook instance, but they remain as separate entity's.

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Rather than use alias' - consider using Distribution Lists. In order to send from a Distribution List, you will need to enable 'Send From' in the security options of the group in Active Directory. Downside of this is that the group can't have 'Send From' rights to itself!

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This video might be useful for new Outlook users who want to know how to create and delete email aliases.

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Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. –  Mark Henderson Sep 27 '12 at 23:24

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