JavaScript (ES6),  83 ... 76  74 bytes

Returns true if \$M\$ does not exist.

Derived from this answer by xnor.

f=(n,q,P=(n,d=n)=>p=--d&&P(n,d)+1-P(n%d?1:d))=>P(n)^q?p>q*q||f(n+1,q||p):n 

Try it online!

How?

Computing \$\phi(n)\$

This is based on the formula:

$$\sum_{d|n}\phi(d)=n$$

which implies:

$$\phi(n)=n-\sum_{d|n,d<n}\phi(d)$$

But in the JS implementation, we actually compute:

$$\begin{align}P(n)&=\sum_{d=1}^{n-1}1-\delta_{d|n}P(d)\\ &=n-1-\sum_{d|n,d<n}P(d)\end{align}$$

It leads to the same results, except \$P(1)=0\$ instead of \$\phi(1)=1\$. This is fine because we don't need to support \$n=1\$, as per the challenge rules. And this allows us to do the following recursive call:

P(n % d ? 1 : d) 

which evaluates to \$0\$ if \$d\$ is not a divisor of \$n\$.

Wrapper code

At each iteration, we compute \$p=P(n)\$. The result of the first iteration is saved into \$q\$. We then increment \$n\$ until \$p=q\$ (success) or \$p>q^2\$ (failure).

Jelly,  11  10 bytes

r²ÆṪẹḢ$+⁸Ḣ 

A monadic Link accepting a positive integer which yields a non-negative integer (0 if no \$M\$ exists).

Try it online!

How?

r²ÆṪẹḢ$+⁸Ḣ - Link: integer, n ² - (n) squared r - (n) inclusive range (n²) ÆṪ - Euler totient (vectorises) $ - last two links as a monad: Ḣ - head - i.e. yield totient(n) and leave [totient(n+1),...,totient(n²)] ẹ - indices of (i.e. a list of offsets to higher Ms) ⁸ - chain's left argument (n) + - add (vectorises) (i.e. a list of higher Ms) Ḣ - head (note head-ing an empty list yields zero) 

Perl 6, 57 52 bytes

-5 bytes using the .& operator thanks to Jo King

{first *.&($!={grep 1,($_ Xgcd^$_)})==.$!,$_^..$_²} 

Try it online!

Returns Nil if no solution was found.

Explanation

{ } # Anonymous block $_ Xgcd^$_ # gcds of m and 0..m-1 grep 1, # Filter 1s { } # Totient function ($!= ) # Assign to $! first ,$_^..$_² # First item of n+1..n² where *.& ==.$! # ϕ(m) == ϕ(n) 

Jelly, 13 12 bytes

r²ÆṪ=€Ḣ$T+⁸Ḣ 

Try it online!

A monadic link taking an integer and returning the next integer with shared totient or zero.

Explanation

Main link (takes integer argument z)

r² | Range from z to z ** 2 inclusive ÆṪ | Totient function of each $ | Following as a monad =€Ḣ | - Check whether each equal to the first, popping the first before doing so T | Truthy indices +⁸ | Plus z Ḣ | - Head (returns 0 if the previous link yielded an empty list) 

05AB1E, 11 10 9 7 bytes

L+.Δ‚ÕË 

-1 byte with help from @ExpiredData.
-2 bytes thanks to @Grimmy.

Outputs -1 if no \$m\$ exists.

Try it online or verify all test cases.

Explanation:

L # Push a list in the range [1, (implicit) input-integer n] + # Add the (implicit) input-integer n to each to make the range [n+1, 2n] .Δ # Get the first value of this list which is truthy for # (or results in -1 if none are truthy): ‚ # Pair the current value with the (implicit) input-integer n Õ # Get the Euler's totient of both Ë # Check whether both are equal to each other # (after which the result is output implicitly) 

Most answers use \$n^2\$ as the range to check in, but this answer uses \$2n\$ instead to save a byte. If we look at the Mathematica implementation on the oeis sequence A066659 we can see it also uses the range \$[n+1, 2n+1)\$ to check in.

Gaia, 11 bytes

:sUt¦ṇ=∆:+¿ 

Try it online!

:s		| push n,n^2 U		| push range [n, n + 1, ..., n^2] t¦		| calculate [phi(n),phi(n+1), ..., phi(n^2)] ṇ		| push [phi(n+1), phi(n+2), ..., phi(n^2)], phi(n) =∆	| find 1-based index of first in the list equal to phi(n), returning 0 if none 	:	| dup the index 	 +¿	| if the index is falsey, do nothing (leaving 0 on the stack) 		| otherwise add (implicitly) n 		| and implicitly print top of stack 

Python 2, 115 bytes

lambda N:next((j for j in range(N+1,max(6,t(N)**2))if t(j)==t(N)),0) t=lambda n:sum(k/n*k%n>n-2for k in range(n*n)) 

Try it online!

Returns 0 for falsey. The totient function t is based on Dennis's answer to a previous question.

Times out on TIO for N=62.

Ruby, 70 68 bytes

->n{(n+1..n*n).find{|m|g=->g{(1..g).sum{|h|1/h.gcd(g)}};g[m]==g[n]}} 

Try it online!

Returns nil if not found.

Python 3, 160 121 bytes

Saved 39 bytes thanks to @JoKing!

Returns None if no \$M\$ exists:

import math t=lambda n:sum(math.gcd(i,n)<2for i in range(n)) def s(x): n=x while n<x*x: n+=1 if t(n)==t(x):return n 

Try it online!

If throwing an exception is allowed when no \$M\$ exists:

Python 3, 145 114 bytes

Saved 31 bytes thanks to @JoKing!

lambda n:[t(i+1)for i in range(n,n*n)].index(t(n))-~n import math t=lambda n:sum(math.gcd(i,n)<2for i in range(n)) 

Try it online!

Python 3, 97 bytes

lambda x,n=1:[n>x*x,x+n][t(x+n)==t(x)]or s(x,n+1) t=lambda n:sum(k//n*k%n>n-2for k in range(n*n)) 

Try it online!

Totient function taken from Chas Brown's answer, originally from Dennis.Returns True for cases where M doesn't exist, though if that doesn't satisfy you, a far less efficient version that returns False is only two bytes longer.

J, 42 31 bytes

(1{]) ::0[([+[:I.{=}.)5 p:i.@*: 

Try it online!

Returns 0 if no \$M\$ exists.

Explanation :

(of the previous version)

The argument is n

 @*: find n^2 and i. make a list 0..n^2-1 + to each number in the list add ] the argument (n) -> list n..n^2+n-1 5 p: find the totient of each number in the above list @( ) use the above result as an argument for ( ) the next verb = compare {. the head of the list ] with each number in the list [:I. get the indices where the above is true ( ) ::0 try the verb in parentheses and return 0 if it failed 1{] get the second (0-based indexing) element + and add n to it 

Japt, 21 19 bytes

ôU ÅæÈo èjX ¥Uo èjU 

Try it

ôU range [input ...input + input] Å cut off first element æÈ return the first element to return a truty value when passed to: o range [0... next element] è number of elements that are.. jX coprime to next element. ¥ equal to Uo èjU num of coprimes of input 

Outputs null if no \$m\$ exists.

Thanks to @Shaggy for reminding me of ô() which gives the range needed